[Script Info]
Title: 
[Events]
Format: Layer, Start, End, Style, Name, MarginL, MarginR, MarginV, Effect, Text
Dialogue: 0,0:00:01.00,0:00:08.42,Default,,0000,0000,0000,,A binomial expression is the\Nsum or difference of two
Dialogue: 0,0:00:08.42,0:00:14.36,Default,,0000,0000,0000,,terms. So for example 2X\Nplus three Y.
Dialogue: 0,0:00:15.00,0:00:17.23,Default,,0000,0000,0000,,Is an example of a binomial
Dialogue: 0,0:00:17.23,0:00:22.31,Default,,0000,0000,0000,,expression. Because it's the sum\Nof the term 2X and the term 3 Y
Dialogue: 0,0:00:22.31,0:00:24.56,Default,,0000,0000,0000,,is the sum of these two terms.
Dialogue: 0,0:00:25.47,0:00:30.29,Default,,0000,0000,0000,,Some of the terms could be just\Nnumbers, so for example X plus
Dialogue: 0,0:00:30.29,0:00:36.22,Default,,0000,0000,0000,,one. Is the sum of the term X\Nand the term one, so that's two
Dialogue: 0,0:00:36.22,0:00:37.67,Default,,0000,0000,0000,,is a binomial expression.
Dialogue: 0,0:00:39.07,0:00:45.89,Default,,0000,0000,0000,,A-B is the difference of\Nthe two terms A&B, so
Dialogue: 0,0:00:45.89,0:00:49.30,Default,,0000,0000,0000,,that too is a binomial
Dialogue: 0,0:00:49.30,0:00:54.01,Default,,0000,0000,0000,,expression. Now in your previous\Nwork you have seen many binomial
Dialogue: 0,0:00:54.01,0:00:57.34,Default,,0000,0000,0000,,expressions and you have raised\Nthem to different powers. So you
Dialogue: 0,0:00:57.34,0:00:59.16,Default,,0000,0000,0000,,have squared them, cube them and
Dialogue: 0,0:00:59.16,0:01:03.88,Default,,0000,0000,0000,,so on. You probably already be\Nvery familiar with working with
Dialogue: 0,0:01:03.88,0:01:07.84,Default,,0000,0000,0000,,the binomial expression like X\NPlus One and squaring it.
Dialogue: 0,0:01:08.52,0:01:13.97,Default,,0000,0000,0000,,And you have done that by\Nremembering that when we want to
Dialogue: 0,0:01:13.97,0:01:17.60,Default,,0000,0000,0000,,square a bracket when\Nmultiplying the bracket by
Dialogue: 0,0:01:17.60,0:01:25.06,Default,,0000,0000,0000,,itself. So X Plus One squared is\NX Plus One multiplied by X plus
Dialogue: 0,0:01:25.06,0:01:29.74,Default,,0000,0000,0000,,one. And we remove the brackets\Nby multiplying all the terms in
Dialogue: 0,0:01:29.74,0:01:33.46,Default,,0000,0000,0000,,the first bracket by all the\Nterms in the SEC bracket, so
Dialogue: 0,0:01:33.46,0:01:35.63,Default,,0000,0000,0000,,they'll be an X multiplied by X.
Dialogue: 0,0:01:35.63,0:01:36.90,Default,,0000,0000,0000,,Which is X squared.
Dialogue: 0,0:01:38.36,0:01:42.85,Default,,0000,0000,0000,,X multiplied by one which\Nis just X.
Dialogue: 0,0:01:44.47,0:01:47.00,Default,,0000,0000,0000,,1 multiplied by X, which is
Dialogue: 0,0:01:47.00,0:01:51.90,Default,,0000,0000,0000,,another X. And 1 * 1, which\Nis just one.
Dialogue: 0,0:01:52.85,0:01:58.67,Default,,0000,0000,0000,,So to tide you all that up X\NPlus One squared is equal to X
Dialogue: 0,0:01:58.67,0:02:04.53,Default,,0000,0000,0000,,squared. As an X plus another X\Nwhich is 2 X.
Dialogue: 0,0:02:04.85,0:02:07.62,Default,,0000,0000,0000,,Plus the one at the end.
Dialogue: 0,0:02:07.62,0:02:11.80,Default,,0000,0000,0000,,Note in particular that we\Nhave two X here and that came
Dialogue: 0,0:02:11.80,0:02:16.32,Default,,0000,0000,0000,,from this X here and another X\Nthere. I'll come back to that
Dialogue: 0,0:02:16.32,0:02:19.45,Default,,0000,0000,0000,,point later on and will see\Nwhy that's important.
Dialogue: 0,0:02:20.64,0:02:24.40,Default,,0000,0000,0000,,Now suppose we want to raise a\Nbinomial expression to our power
Dialogue: 0,0:02:24.40,0:02:27.84,Default,,0000,0000,0000,,that's higher than two. So\Nsuppose we want to cube it,
Dialogue: 0,0:02:27.84,0:02:31.28,Default,,0000,0000,0000,,raise it to the power four or\Nfive or even 32.
Dialogue: 0,0:02:31.82,0:02:34.78,Default,,0000,0000,0000,,The process of removing the\Nbrackets by multiplying term by
Dialogue: 0,0:02:34.78,0:02:38.63,Default,,0000,0000,0000,,term over and over again is very\Nvery cumbersome. I mean, if we
Dialogue: 0,0:02:38.63,0:02:42.18,Default,,0000,0000,0000,,wanted to workout X plus one to\Nthe Seven, you wouldn't really
Dialogue: 0,0:02:42.18,0:02:45.14,Default,,0000,0000,0000,,want to multiply a pair of\Nbrackets by itself several
Dialogue: 0,0:02:45.14,0:02:48.99,Default,,0000,0000,0000,,times. So what we want is a\Nbetter way. Better way of doing
Dialogue: 0,0:02:48.99,0:02:54.07,Default,,0000,0000,0000,,that. And one way of doing it\Nis by means of a triangle of
Dialogue: 0,0:02:54.07,0:02:55.95,Default,,0000,0000,0000,,numbers, which is called\NPascals Triangle.
Dialogue: 0,0:02:57.01,0:03:01.22,Default,,0000,0000,0000,,Pascal was a 17th century French\Nmathematician and he derived
Dialogue: 0,0:03:01.22,0:03:05.43,Default,,0000,0000,0000,,this triangle of numbers that\Nwill repeat for ourselves now,
Dialogue: 0,0:03:05.43,0:03:10.48,Default,,0000,0000,0000,,and this is how we form the\Ntriangle. We start by writing
Dialogue: 0,0:03:10.48,0:03:12.17,Default,,0000,0000,0000,,down the number one.
Dialogue: 0,0:03:14.26,0:03:20.13,Default,,0000,0000,0000,,Then we form a new row and on\Nthis nuro we have a one.
Dialogue: 0,0:03:20.63,0:03:21.91,Default,,0000,0000,0000,,And another one.
Dialogue: 0,0:03:23.73,0:03:27.73,Default,,0000,0000,0000,,We're going to build up a\Ntriangle like this and each nuro
Dialogue: 0,0:03:27.73,0:03:29.72,Default,,0000,0000,0000,,that we write down will start
Dialogue: 0,0:03:29.72,0:03:32.44,Default,,0000,0000,0000,,with a one. And will end with a
Dialogue: 0,0:03:32.44,0:03:36.52,Default,,0000,0000,0000,,one. So my third row is\Ngoing to begin with the
Dialogue: 0,0:03:36.52,0:03:38.32,Default,,0000,0000,0000,,one and end with a one.
Dialogue: 0,0:03:39.49,0:03:41.44,Default,,0000,0000,0000,,And in a few minutes,\Nwe'll write a number
Dialogue: 0,0:03:41.44,0:03:42.53,Default,,0000,0000,0000,,in there in the gap.
Dialogue: 0,0:03:43.81,0:03:47.84,Default,,0000,0000,0000,,The next row will begin with a\None and end with a one and will
Dialogue: 0,0:03:47.84,0:03:49.73,Default,,0000,0000,0000,,write a number in there and a
Dialogue: 0,0:03:49.73,0:03:53.94,Default,,0000,0000,0000,,number in there. And in this way\Nwe can build a triangle of
Dialogue: 0,0:03:53.94,0:03:56.93,Default,,0000,0000,0000,,numbers and we can build it as\Nbig as we want to.
Dialogue: 0,0:03:58.03,0:04:00.04,Default,,0000,0000,0000,,How do we find this number in
Dialogue: 0,0:04:00.04,0:04:03.77,Default,,0000,0000,0000,,here? Well, the number that\Ngoes in here we find by
Dialogue: 0,0:04:03.77,0:04:05.16,Default,,0000,0000,0000,,looking on the row above.
Dialogue: 0,0:04:06.46,0:04:10.23,Default,,0000,0000,0000,,And looking above to the left\Nand above to the right.
Dialogue: 0,0:04:11.32,0:04:15.70,Default,,0000,0000,0000,,And adding what we find, there's\Na one here. There's a one there.
Dialogue: 0,0:04:15.70,0:04:20.42,Default,,0000,0000,0000,,We add them one and one gives 2\Nand we write the result in
Dialogue: 0,0:04:20.42,0:04:21.77,Default,,0000,0000,0000,,there. So there's two.
Dialogue: 0,0:04:22.31,0:04:27.66,Default,,0000,0000,0000,,On the 3rd row has come by\Nadding that one and that
Dialogue: 0,0:04:27.66,0:04:28.55,Default,,0000,0000,0000,,one together.
Dialogue: 0,0:04:30.22,0:04:34.52,Default,,0000,0000,0000,,Let's look at the next row down\Nthe number that's going to go in
Dialogue: 0,0:04:34.52,0:04:38.28,Default,,0000,0000,0000,,here. Is found by looking\Non the previous row.
Dialogue: 0,0:04:39.91,0:04:43.74,Default,,0000,0000,0000,,And we look above left which\Ngives us the one we look above
Dialogue: 0,0:04:43.74,0:04:47.28,Default,,0000,0000,0000,,to the right, which gives us\Ntwo, and we add the numbers
Dialogue: 0,0:04:47.28,0:04:51.42,Default,,0000,0000,0000,,that we find, so we're adding\Na one and two which is 3 and
Dialogue: 0,0:04:51.42,0:04:52.89,Default,,0000,0000,0000,,we write that in there.
Dialogue: 0,0:04:54.21,0:04:55.74,Default,,0000,0000,0000,,What about the number here?
Dialogue: 0,0:04:57.92,0:05:02.04,Default,,0000,0000,0000,,Well again previous row above\Nto the left is 2 above to the
Dialogue: 0,0:05:02.04,0:05:06.80,Default,,0000,0000,0000,,right is one. We add what we\Nfind 2 plus one is 3 and that
Dialogue: 0,0:05:06.80,0:05:07.75,Default,,0000,0000,0000,,goes in there.
Dialogue: 0,0:05:08.84,0:05:12.09,Default,,0000,0000,0000,,And we can carry on building\Nthis triangle as big as we want
Dialogue: 0,0:05:12.09,0:05:13.84,Default,,0000,0000,0000,,to. Let's just do one more row.
Dialogue: 0,0:05:14.37,0:05:16.03,Default,,0000,0000,0000,,We start the row.
Dialogue: 0,0:05:16.82,0:05:20.94,Default,,0000,0000,0000,,With a one and we finished with\Na one and we put some numbers in
Dialogue: 0,0:05:20.94,0:05:22.87,Default,,0000,0000,0000,,here and in here and in here.
Dialogue: 0,0:05:24.27,0:05:26.17,Default,,0000,0000,0000,,The number that's going\Nto go in here.
Dialogue: 0,0:05:27.26,0:05:31.50,Default,,0000,0000,0000,,Is found from the previous row\Nby adding the one and the three.
Dialogue: 0,0:05:31.50,0:05:32.80,Default,,0000,0000,0000,,So 1 + 3.
Dialogue: 0,0:05:33.34,0:05:35.63,Default,,0000,0000,0000,,Is 4 let me write that in there.
Dialogue: 0,0:05:37.05,0:05:38.89,Default,,0000,0000,0000,,The number that's going to go in
Dialogue: 0,0:05:38.89,0:05:42.32,Default,,0000,0000,0000,,here. Well, we look in the\Nprevious row above left and
Dialogue: 0,0:05:42.32,0:05:46.49,Default,,0000,0000,0000,,above right. 3 + 3 is 6 and we\Nwrite that in there.
Dialogue: 0,0:05:48.50,0:05:53.26,Default,,0000,0000,0000,,And finally 3 Plus One is 4 when\Nwe write that in there. So
Dialogue: 0,0:05:53.26,0:05:54.28,Default,,0000,0000,0000,,that's another row.
Dialogue: 0,0:05:54.92,0:05:59.37,Default,,0000,0000,0000,,And what you should do now is\Npractice generating additional
Dialogue: 0,0:05:59.37,0:06:02.93,Default,,0000,0000,0000,,rows for yourself, and\Naltogether this triangle of
Dialogue: 0,0:06:02.93,0:06:04.71,Default,,0000,0000,0000,,numbers is called pascals.
Dialogue: 0,0:06:05.81,0:06:08.95,Default,,0000,0000,0000,,Pascal's triangle.
Dialogue: 0,0:06:09.06,0:06:15.79,Default,,0000,0000,0000,,OK.\NNow we're going to use this
Dialogue: 0,0:06:15.79,0:06:19.69,Default,,0000,0000,0000,,triangle to expand binomial\Nexpressions and will see that it
Dialogue: 0,0:06:19.69,0:06:22.42,Default,,0000,0000,0000,,can make life very easy for us.
Dialogue: 0,0:06:22.44,0:06:28.66,Default,,0000,0000,0000,,We'll start by\Ngoing back to
Dialogue: 0,0:06:28.66,0:06:31.77,Default,,0000,0000,0000,,the expression A+B.
Dialogue: 0,0:06:31.88,0:06:33.43,Default,,0000,0000,0000,,To the power 2.
Dialogue: 0,0:06:34.15,0:06:37.45,Default,,0000,0000,0000,,So we have binomial binomial\Nexpression here, a I'd be and
Dialogue: 0,0:06:37.45,0:06:39.55,Default,,0000,0000,0000,,we're raising it to the power 2.
Dialogue: 0,0:06:40.13,0:06:47.05,Default,,0000,0000,0000,,Let's do it the old way. First\Nof all by multiplying A&B by
Dialogue: 0,0:06:47.05,0:06:50.88,Default,,0000,0000,0000,,itself. Because we're squaring
Dialogue: 0,0:06:50.88,0:06:55.04,Default,,0000,0000,0000,,A&B. Let's write down\Nwhat will get.
Dialogue: 0,0:06:56.57,0:07:00.47,Default,,0000,0000,0000,,A multiplied by a\Ngives us a squared.
Dialogue: 0,0:07:02.63,0:07:10.26,Default,,0000,0000,0000,,A multiplied by B will give us\Na Times B or just a B.
Dialogue: 0,0:07:10.37,0:07:17.30,Default,,0000,0000,0000,,Be multiplied by a.\NGives us a BA.
Dialogue: 0,0:07:18.22,0:07:23.18,Default,,0000,0000,0000,,And finally, be multiplied by B.\NGive us a B squared.
Dialogue: 0,0:07:25.00,0:07:29.79,Default,,0000,0000,0000,,And if we just tidy it, what we\Nfound, there's a squared.
Dialogue: 0,0:07:30.48,0:07:31.54,Default,,0000,0000,0000,,There's an AB.
Dialogue: 0,0:07:32.08,0:07:37.19,Default,,0000,0000,0000,,And because BA is the same as a\Nbee, there's another a be here.
Dialogue: 0,0:07:37.19,0:07:40.11,Default,,0000,0000,0000,,So altogether there's two lots\Nof a B.
Dialogue: 0,0:07:40.13,0:07:45.90,Default,,0000,0000,0000,,And finally, AB squared\Nat the end.
Dialogue: 0,0:07:46.40,0:07:49.58,Default,,0000,0000,0000,,Now that's the sort of\Nexpansion. This sort of removing
Dialogue: 0,0:07:49.58,0:07:52.76,Default,,0000,0000,0000,,brackets that you've seen many\Ntimes before. You were already
Dialogue: 0,0:07:52.76,0:07:56.58,Default,,0000,0000,0000,,very familiar with, but what I\Nwant to do is make some
Dialogue: 0,0:07:56.58,0:07:57.85,Default,,0000,0000,0000,,observations about this result.
Dialogue: 0,0:07:58.89,0:08:04.26,Default,,0000,0000,0000,,When we expanded A+B to the\Npower two, what we find is that
Dialogue: 0,0:08:04.26,0:08:08.39,Default,,0000,0000,0000,,as we successively move through\Nthese terms that we've written
Dialogue: 0,0:08:08.39,0:08:13.76,Default,,0000,0000,0000,,down the power of a decreases,\Nit starts off here with an A
Dialogue: 0,0:08:13.76,0:08:17.91,Default,,0000,0000,0000,,squared. The highest power being\Ntwo corresponding to the power
Dialogue: 0,0:08:17.91,0:08:20.64,Default,,0000,0000,0000,,in the original binomial\Nexpression, and then every
Dialogue: 0,0:08:20.64,0:08:24.75,Default,,0000,0000,0000,,subsequent term that power\Ndrops. So it was 8 to the power
Dialogue: 0,0:08:24.75,0:08:29.15,Default,,0000,0000,0000,,2. There's A to the power one\Nin here, although we don't
Dialogue: 0,0:08:29.15,0:08:33.30,Default,,0000,0000,0000,,normally right the one in and\Nthen know as at all, so the
Dialogue: 0,0:08:33.30,0:08:36.80,Default,,0000,0000,0000,,powers of a decrease as we\Nmove from left to right.
Dialogue: 0,0:08:38.40,0:08:39.51,Default,,0000,0000,0000,,What about bees?
Dialogue: 0,0:08:40.11,0:08:43.88,Default,,0000,0000,0000,,There's no bees in here.\NThere's a beta. The one in
Dialogue: 0,0:08:43.88,0:08:46.97,Default,,0000,0000,0000,,there, although we just\Nnormally right B&AB to the
Dialogue: 0,0:08:46.97,0:08:51.43,Default,,0000,0000,0000,,power two there. So as we move\Nfrom left to right, the powers
Dialogue: 0,0:08:51.43,0:08:55.20,Default,,0000,0000,0000,,of B increase until we reach\Nthe highest power B squared,
Dialogue: 0,0:08:55.20,0:08:58.63,Default,,0000,0000,0000,,and the squared corresponds to\Nthe two in the original
Dialogue: 0,0:08:58.63,0:08:58.98,Default,,0000,0000,0000,,problem.
Dialogue: 0,0:09:00.55,0:09:03.99,Default,,0000,0000,0000,,What else can we observe if we\Nlook at the coefficients of
Dialogue: 0,0:09:03.99,0:09:07.44,Default,,0000,0000,0000,,these terms now the coefficients\Nare the numbers in front of each
Dialogue: 0,0:09:07.44,0:09:10.60,Default,,0000,0000,0000,,of these terms. Well, there's a\None in here, although we
Dialogue: 0,0:09:10.60,0:09:13.75,Default,,0000,0000,0000,,wouldn't normally write it in,\Nthere's a two there, and there's
Dialogue: 0,0:09:13.75,0:09:16.62,Default,,0000,0000,0000,,a one inference of the B\Nsquared, although we wouldn't
Dialogue: 0,0:09:16.62,0:09:17.77,Default,,0000,0000,0000,,normally write it in.
Dialogue: 0,0:09:19.21,0:09:21.89,Default,,0000,0000,0000,,So the coefficients\Nare 1, two and one.
Dialogue: 0,0:09:23.05,0:09:27.13,Default,,0000,0000,0000,,Now let me remind you again\Nabout pascals triangle. Have a
Dialogue: 0,0:09:27.13,0:09:32.70,Default,,0000,0000,0000,,copy of the triangle here so we\Ncan refer to it. If we look at
Dialogue: 0,0:09:32.70,0:09:37.15,Default,,0000,0000,0000,,pascals triangle here will see\Nthat one 2 one is the numbers
Dialogue: 0,0:09:37.15,0:09:41.60,Default,,0000,0000,0000,,that's in the 3rd row of pascals\Ntriangle. 121 other numbers that
Dialogue: 0,0:09:41.60,0:09:45.31,Default,,0000,0000,0000,,occur in the expansion of A+B to\Nthe power 2.
Dialogue: 0,0:09:46.55,0:09:51.13,Default,,0000,0000,0000,,There's something else I want to\Npoint out that this 2A. B in
Dialogue: 0,0:09:51.13,0:09:56.69,Default,,0000,0000,0000,,here. Came from a term here 1A B\Non one BA in there and together
Dialogue: 0,0:09:56.69,0:10:01.36,Default,,0000,0000,0000,,the one plus the one gave the\Ntwo in exactly the same way as
Dialogue: 0,0:10:01.36,0:10:05.70,Default,,0000,0000,0000,,the two in pascals triangle came\Nfrom adding the one and the one
Dialogue: 0,0:10:05.70,0:10:07.04,Default,,0000,0000,0000,,in the previous row.
Dialogue: 0,0:10:08.14,0:10:12.47,Default,,0000,0000,0000,,So Pascal's triangle will give\Nus an easy way of evaluating a
Dialogue: 0,0:10:12.47,0:10:16.80,Default,,0000,0000,0000,,binomial expression when we want\Nto raise it to an even higher
Dialogue: 0,0:10:16.80,0:10:21.86,Default,,0000,0000,0000,,power. Let me look at what\Nhappens if we want a plus B to
Dialogue: 0,0:10:21.86,0:10:26.19,Default,,0000,0000,0000,,the power three and will see\Nthat we can do this almost
Dialogue: 0,0:10:26.19,0:10:31.04,Default,,0000,0000,0000,,straight away. What we note is\Nthat the highest power now is 3.
Dialogue: 0,0:10:31.61,0:10:34.33,Default,,0000,0000,0000,,So we start with an\NA to the power 3.
Dialogue: 0,0:10:35.46,0:10:41.62,Default,,0000,0000,0000,,Each successive term that power\Nof a will reduce, so they'll be
Dialogue: 0,0:10:41.62,0:10:44.18,Default,,0000,0000,0000,,a term in a squared.
Dialogue: 0,0:10:45.12,0:10:46.67,Default,,0000,0000,0000,,That'll be a term in A.
Dialogue: 0,0:10:47.46,0:10:50.00,Default,,0000,0000,0000,,And then they'll be a term\Nwithout any Asian at all.
Dialogue: 0,0:10:50.82,0:10:55.93,Default,,0000,0000,0000,,So as we move from left to\Nright, the powers of a decrease.
Dialogue: 0,0:10:58.47,0:11:01.85,Default,,0000,0000,0000,,Similarly, as we move from left\Nto right, we want the powers of
Dialogue: 0,0:11:01.85,0:11:05.49,Default,,0000,0000,0000,,be to increase just as they did\Nhere. There will be no bees in
Dialogue: 0,0:11:05.49,0:11:09.57,Default,,0000,0000,0000,,the first term. ABB to the power\None or just B.
Dialogue: 0,0:11:10.30,0:11:11.42,Default,,0000,0000,0000,,In the second term.
Dialogue: 0,0:11:12.15,0:11:14.11,Default,,0000,0000,0000,,B to the power two in the next
Dialogue: 0,0:11:14.11,0:11:18.56,Default,,0000,0000,0000,,term. And then finally there\Nwill be a B to the power
Dialogue: 0,0:11:18.56,0:11:22.58,Default,,0000,0000,0000,,three and we stop it be to\Nthe power three that highest
Dialogue: 0,0:11:22.58,0:11:25.26,Default,,0000,0000,0000,,power corresponding to the\Npower in the original
Dialogue: 0,0:11:25.26,0:11:25.92,Default,,0000,0000,0000,,binomial expression.
Dialogue: 0,0:11:27.92,0:11:31.16,Default,,0000,0000,0000,,We need some coefficients.\NThat's the numbers in front of
Dialogue: 0,0:11:31.16,0:11:32.46,Default,,0000,0000,0000,,each of these terms.
Dialogue: 0,0:11:33.35,0:11:36.75,Default,,0000,0000,0000,,And the numbers come from the\Nrelevant row in pascals
Dialogue: 0,0:11:36.75,0:11:40.83,Default,,0000,0000,0000,,triangle, and we want the row\Nthat begins 1, three and the
Dialogue: 0,0:11:40.83,0:11:44.57,Default,,0000,0000,0000,,reason why we want the row\Nbeginning 1. Three is because
Dialogue: 0,0:11:44.57,0:11:48.31,Default,,0000,0000,0000,,three is the power in the\Noriginal expression. So I go
Dialogue: 0,0:11:48.31,0:11:52.73,Default,,0000,0000,0000,,back to my pascals triangle and\NI look for the robe beginning 1
Dialogue: 0,0:11:52.73,0:11:54.09,Default,,0000,0000,0000,,three, which is 1331.
Dialogue: 0,0:11:55.12,0:11:58.14,Default,,0000,0000,0000,,So these numbers are the\Ncoefficients that I need.
Dialogue: 0,0:11:58.71,0:12:01.44,Default,,0000,0000,0000,,In this expansion I want one.
Dialogue: 0,0:12:02.39,0:12:10.03,Default,,0000,0000,0000,,331\NAnd just to tidy that up a
Dialogue: 0,0:12:10.03,0:12:13.15,Default,,0000,0000,0000,,little bit 1A Cube would\Nnormally just be written as a
Dialogue: 0,0:12:13.15,0:12:16.61,Default,,0000,0000,0000,,cubed. 3A squared
Dialogue: 0,0:12:16.61,0:12:19.82,Default,,0000,0000,0000,,B. 3A B
Dialogue: 0,0:12:19.82,0:12:25.46,Default,,0000,0000,0000,,squared. And finally 1B cubed\Nwhich would normally write as
Dialogue: 0,0:12:25.46,0:12:27.01,Default,,0000,0000,0000,,just be cubed.
Dialogue: 0,0:12:27.57,0:12:33.58,Default,,0000,0000,0000,,Now, I hope you'll agree that\Nusing pascals triangle to expand
Dialogue: 0,0:12:33.58,0:12:36.31,Default,,0000,0000,0000,,A+B to the power 3.
Dialogue: 0,0:12:36.97,0:12:43.17,Default,,0000,0000,0000,,Is much simpler than multiplying\NA+B Times A+B times A+B?
Dialogue: 0,0:12:43.67,0:12:47.25,Default,,0000,0000,0000,,What I want to do for just\Nbefore we go on is just actually
Dialogue: 0,0:12:47.25,0:12:50.33,Default,,0000,0000,0000,,go and do it the long way, just\Nto point something out.
Dialogue: 0,0:12:50.33,0:12:56.03,Default,,0000,0000,0000,,Let's go back to\Na plus B.
Dialogue: 0,0:12:56.05,0:13:01.37,Default,,0000,0000,0000,,To the power three and work\Nit out the long way by noting
Dialogue: 0,0:13:01.37,0:13:07.09,Default,,0000,0000,0000,,that we can work this out as\Na plus B multiplied by a plus
Dialogue: 0,0:13:07.09,0:13:08.32,Default,,0000,0000,0000,,B or squared.
Dialogue: 0,0:13:09.89,0:13:14.91,Default,,0000,0000,0000,,We've already expanded A+B to\Nthe power two, so let's write
Dialogue: 0,0:13:14.91,0:13:22.47,Default,,0000,0000,0000,,that down. Well remember A+B to\Nthe power two we've already seen
Dialogue: 0,0:13:22.47,0:13:24.43,Default,,0000,0000,0000,,is A squared.
Dialogue: 0,0:13:24.43,0:13:29.41,Default,,0000,0000,0000,,2-AB And\NB squared.
Dialogue: 0,0:13:31.71,0:13:34.85,Default,,0000,0000,0000,,Now to expand this,\Neverything in the first
Dialogue: 0,0:13:34.85,0:13:37.60,Default,,0000,0000,0000,,bracket must multiply\Neverything in the SEC
Dialogue: 0,0:13:37.60,0:13:41.14,Default,,0000,0000,0000,,bracket, so we've been a\Nmultiplied by a squared
Dialogue: 0,0:13:41.14,0:13:42.71,Default,,0000,0000,0000,,which is a cubed.
Dialogue: 0,0:13:45.33,0:13:47.58,Default,,0000,0000,0000,,A multiplied by two AB.
Dialogue: 0,0:13:48.37,0:13:49.44,Default,,0000,0000,0000,,Which is 2.
Dialogue: 0,0:13:50.00,0:13:52.80,Default,,0000,0000,0000,,A squared B.
Dialogue: 0,0:13:53.08,0:13:56.53,Default,,0000,0000,0000,,A multiplied by
Dialogue: 0,0:13:56.53,0:14:02.47,Default,,0000,0000,0000,,B squared. Which\Nis a B squared.
Dialogue: 0,0:14:04.97,0:14:08.94,Default,,0000,0000,0000,,We be multiplied by a squared.
Dialogue: 0,0:14:08.94,0:14:10.64,Default,,0000,0000,0000,,She's BA squared.
Dialogue: 0,0:14:12.39,0:14:18.59,Default,,0000,0000,0000,,We be multiplied by two AB which\Nis 2A B squared.
Dialogue: 0,0:14:18.69,0:14:24.76,Default,,0000,0000,0000,,And finally, be multiplied by AB\Nsquared is AB cubed.
Dialogue: 0,0:14:25.34,0:14:33.15,Default,,0000,0000,0000,,To tidy this up as a\Ncubed and then notice there's a
Dialogue: 0,0:14:33.15,0:14:36.41,Default,,0000,0000,0000,,squared B terms in here.
Dialogue: 0,0:14:37.52,0:14:41.89,Default,,0000,0000,0000,,And there's also an A squared B\Nturn there, one of them, so
Dialogue: 0,0:14:41.89,0:14:46.59,Default,,0000,0000,0000,,we've into there and a one there\Ntoo, and the one gives you three
Dialogue: 0,0:14:46.59,0:14:48.27,Default,,0000,0000,0000,,lots of A squared fee.
Dialogue: 0,0:14:50.67,0:14:52.66,Default,,0000,0000,0000,,There's an AB squared.
Dialogue: 0,0:14:53.37,0:14:57.27,Default,,0000,0000,0000,,Here, and there's more AB\Nsquared's there. There's one
Dialogue: 0,0:14:57.27,0:15:02.03,Default,,0000,0000,0000,,there, two of them there so\Naltogether will have three lots
Dialogue: 0,0:15:02.03,0:15:03.33,Default,,0000,0000,0000,,of AB squared.
Dialogue: 0,0:15:04.57,0:15:07.79,Default,,0000,0000,0000,,And finally, the last term at\Nthe end B cubed.
Dialogue: 0,0:15:08.60,0:15:12.21,Default,,0000,0000,0000,,That's working out the expansion\Nthe long way. Why have I done
Dialogue: 0,0:15:12.21,0:15:15.82,Default,,0000,0000,0000,,that? Well, I've only done that\Njust to point out something to
Dialogue: 0,0:15:15.82,0:15:20.64,Default,,0000,0000,0000,,you and I want to point out that\Nthe three in here in the three A
Dialogue: 0,0:15:20.64,0:15:22.75,Default,,0000,0000,0000,,squared B came from adding a 2
Dialogue: 0,0:15:22.75,0:15:28.38,Default,,0000,0000,0000,,here. And a one in there 2 plus\Nthe one gave you the three.
Dialogue: 0,0:15:29.18,0:15:33.19,Default,,0000,0000,0000,,Similarly, this three here came\Nfrom a one lot of AB squared
Dialogue: 0,0:15:33.19,0:15:37.86,Default,,0000,0000,0000,,there and two lots of AB squared\Nthere. So the one plus the two
Dialogue: 0,0:15:37.86,0:15:41.87,Default,,0000,0000,0000,,gave you the three, and that\Nmirrors exactly what we had when
Dialogue: 0,0:15:41.87,0:15:45.21,Default,,0000,0000,0000,,we generated the triangle,\Nbecause the three here came back
Dialogue: 0,0:15:45.21,0:15:49.89,Default,,0000,0000,0000,,from adding the one in the two\Nin the row above and the three
Dialogue: 0,0:15:49.89,0:15:53.56,Default,,0000,0000,0000,,here came from adding two and\None in the row above.
Dialogue: 0,0:15:53.62,0:16:00.09,Default,,0000,0000,0000,,Let's have a look at another\Nexample and see if we can just
Dialogue: 0,0:16:00.09,0:16:04.58,Default,,0000,0000,0000,,write the answer down\Nstraightaway. Suppose we want to
Dialogue: 0,0:16:04.58,0:16:07.56,Default,,0000,0000,0000,,expand A+B or raised to the
Dialogue: 0,0:16:07.56,0:16:12.83,Default,,0000,0000,0000,,power 4. Well, this is\Nstraightforward to do. We know
Dialogue: 0,0:16:12.83,0:16:18.17,Default,,0000,0000,0000,,that when we expand this, our\Nhighest power of a will be 4
Dialogue: 0,0:16:18.17,0:16:21.46,Default,,0000,0000,0000,,because that's the power in\Nthe original expression.
Dialogue: 0,0:16:22.65,0:16:26.80,Default,,0000,0000,0000,,And thereafter every subsequent\Nterm will have a power reduced
Dialogue: 0,0:16:26.80,0:16:31.36,Default,,0000,0000,0000,,by one each time. So there will\Nbe an A cubed.
Dialogue: 0,0:16:32.28,0:16:36.68,Default,,0000,0000,0000,,And a squared and A and then, no\Nworries at all.
Dialogue: 0,0:16:38.09,0:16:41.31,Default,,0000,0000,0000,,As we move from left to\Nright, the powers of B will
Dialogue: 0,0:16:41.31,0:16:44.25,Default,,0000,0000,0000,,increase. There will be none\Nat all in the first term.
Dialogue: 0,0:16:45.47,0:16:47.40,Default,,0000,0000,0000,,And they'll be a big to the one.
Dialogue: 0,0:16:47.40,0:16:50.01,Default,,0000,0000,0000,,Or just be. A bit of the two.
Dialogue: 0,0:16:51.25,0:16:55.90,Default,,0000,0000,0000,,Beta three will be cubed and\Nfinally the last term will be to
Dialogue: 0,0:16:55.90,0:16:59.84,Default,,0000,0000,0000,,the four and again the highest\Npower corresponding to the power
Dialogue: 0,0:16:59.84,0:17:01.63,Default,,0000,0000,0000,,four in the original expression.
Dialogue: 0,0:17:02.85,0:17:06.60,Default,,0000,0000,0000,,And all we need now are the\Ncoefficients. The coefficients
Dialogue: 0,0:17:06.60,0:17:10.72,Default,,0000,0000,0000,,come from the appropriate role\Nin the triangle and this time
Dialogue: 0,0:17:10.72,0:17:15.60,Default,,0000,0000,0000,,because we're looking at power\Nfour, we want to look at the Roo
Dialogue: 0,0:17:15.60,0:17:21.57,Default,,0000,0000,0000,,beginning 14. The row beginning\N1 four is 14641.
Dialogue: 0,0:17:22.22,0:17:23.68,Default,,0000,0000,0000,,Those are the coefficients that
Dialogue: 0,0:17:23.68,0:17:30.58,Default,,0000,0000,0000,,will need.\N14641
Dialogue: 0,0:17:31.98,0:17:37.74,Default,,0000,0000,0000,,And just to tidy it up, we\Nwouldn't normally right the one
Dialogue: 0,0:17:37.74,0:17:45.42,Default,,0000,0000,0000,,in there and the one in there so\NA&B to the four is 8 to 4
Dialogue: 0,0:17:45.42,0:17:48.30,Default,,0000,0000,0000,,four A cubed B that's that.
Dialogue: 0,0:17:48.85,0:17:52.14,Default,,0000,0000,0000,,6A squared, B squared.
Dialogue: 0,0:17:52.14,0:17:55.57,Default,,0000,0000,0000,,4A B
Dialogue: 0,0:17:55.57,0:18:02.41,Default,,0000,0000,0000,,cubed. And finally, be\Nto the power 4.
Dialogue: 0,0:18:03.82,0:18:07.67,Default,,0000,0000,0000,,OK, so I hope you'll agree that\Nusing pascals triangle to get
Dialogue: 0,0:18:07.67,0:18:10.56,Default,,0000,0000,0000,,this expansion was much simpler\Nthan multiplying this bracket
Dialogue: 0,0:18:10.56,0:18:14.73,Default,,0000,0000,0000,,over and over by itself. Lots\Nand lots of times that way is
Dialogue: 0,0:18:14.73,0:18:18.91,Default,,0000,0000,0000,,also prone to error, so if you\Ncan get used to using pascals
Dialogue: 0,0:18:18.91,0:18:24.07,Default,,0000,0000,0000,,triangle. We can use the same\Ntechnique even when we have
Dialogue: 0,0:18:24.07,0:18:27.10,Default,,0000,0000,0000,,slightly more complicated\Nexpressions. Let's do another
Dialogue: 0,0:18:27.10,0:18:34.74,Default,,0000,0000,0000,,example. Suppose we want to\Nexpand 2X plus Y all to
Dialogue: 0,0:18:34.74,0:18:36.70,Default,,0000,0000,0000,,the power 3.
Dialogue: 0,0:18:37.84,0:18:40.68,Default,,0000,0000,0000,,So it's more complicated this\Ntime because I just haven't got
Dialogue: 0,0:18:40.68,0:18:43.77,Default,,0000,0000,0000,,a single term here, but I've\Nactually got a 2X in there.
Dialogue: 0,0:18:45.07,0:18:48.07,Default,,0000,0000,0000,,The principle is\Nexactly the same.
Dialogue: 0,0:18:49.51,0:18:53.36,Default,,0000,0000,0000,,What will do is will write this\Nterm down first. The whole of
Dialogue: 0,0:18:53.36,0:18:58.16,Default,,0000,0000,0000,,2X. And just like before, it\Nwill be raised to the highest
Dialogue: 0,0:18:58.16,0:19:02.21,Default,,0000,0000,0000,,possible power which is 3 and\Nthat corresponds to the three in
Dialogue: 0,0:19:02.21,0:19:03.23,Default,,0000,0000,0000,,the original problem.
Dialogue: 0,0:19:06.14,0:19:11.14,Default,,0000,0000,0000,,Every subsequent term will have\Na 2X in it, but as we go from
Dialogue: 0,0:19:11.14,0:19:15.78,Default,,0000,0000,0000,,left to right, the power of 2X\Nwill decrease, so the next term
Dialogue: 0,0:19:15.78,0:19:17.92,Default,,0000,0000,0000,,will have a 2X or squared.
Dialogue: 0,0:19:19.25,0:19:23.68,Default,,0000,0000,0000,,The next term will have a 2X to\Nthe power one or just 2X, and
Dialogue: 0,0:19:23.68,0:19:26.92,Default,,0000,0000,0000,,then there won't be any at all\Nin the last term.
Dialogue: 0,0:19:29.95,0:19:34.30,Default,,0000,0000,0000,,Powers of Y will increase as we\Nmove from the left to the right,
Dialogue: 0,0:19:34.30,0:19:36.48,Default,,0000,0000,0000,,so there won't be any in the
Dialogue: 0,0:19:36.48,0:19:38.78,Default,,0000,0000,0000,,first term. Then they'll be Y.
Dialogue: 0,0:19:39.60,0:19:43.35,Default,,0000,0000,0000,,Then they'll be Y squared and\Nfinally Y cubed.
Dialogue: 0,0:19:45.83,0:19:49.36,Default,,0000,0000,0000,,And then we remember the\Ncoefficients. Where do we get
Dialogue: 0,0:19:49.36,0:19:50.42,Default,,0000,0000,0000,,the coefficients from?
Dialogue: 0,0:19:50.99,0:19:54.71,Default,,0000,0000,0000,,Well, because we're looking at\Npower three, we go to pascals
Dialogue: 0,0:19:54.71,0:19:57.07,Default,,0000,0000,0000,,triangle and we look for the row
Dialogue: 0,0:19:57.07,0:20:01.33,Default,,0000,0000,0000,,beginning 13. You might even\Nremember those numbers now.
Dialogue: 0,0:20:01.33,0:20:06.66,Default,,0000,0000,0000,,We've seen it so many times. The\Nnumbers are 1331. Those are the
Dialogue: 0,0:20:06.66,0:20:08.30,Default,,0000,0000,0000,,coefficients we require, 1331.
Dialogue: 0,0:20:09.01,0:20:13.50,Default,,0000,0000,0000,,So I want one of those\Nthree of those three of
Dialogue: 0,0:20:13.50,0:20:15.13,Default,,0000,0000,0000,,those, one of those.
Dialogue: 0,0:20:16.79,0:20:20.55,Default,,0000,0000,0000,,And there's just a bit\Nmore tidying up to do to
Dialogue: 0,0:20:20.55,0:20:21.58,Default,,0000,0000,0000,,finish it off.
Dialogue: 0,0:20:22.76,0:20:26.48,Default,,0000,0000,0000,,Here we've got 2 to the Power 3,\Ntwo cubed that's eight.
Dialogue: 0,0:20:27.23,0:20:33.02,Default,,0000,0000,0000,,X cubed\NAnd the one just is, one could
Dialogue: 0,0:20:33.02,0:20:36.22,Default,,0000,0000,0000,,just stay there 1. Multiply by\Nall that is not going to do
Dialogue: 0,0:20:36.22,0:20:37.45,Default,,0000,0000,0000,,anything else, just 8X cubed.
Dialogue: 0,0:20:38.39,0:20:42.85,Default,,0000,0000,0000,,What about this term? There's a\N2 squared, which is 4, and it's
Dialogue: 0,0:20:42.85,0:20:47.65,Default,,0000,0000,0000,,got to be multiplied by three.\NSo 4 threes are 12, so we have
Dialogue: 0,0:20:47.65,0:20:53.54,Default,,0000,0000,0000,,12. What about powers of X?\NWell, there be an X squared.
Dialogue: 0,0:20:53.64,0:20:54.83,Default,,0000,0000,0000,,Why?
Dialogue: 0,0:20:56.22,0:21:00.83,Default,,0000,0000,0000,,In this term, we've just got\N2X to the power one. That's
Dialogue: 0,0:21:00.83,0:21:05.44,Default,,0000,0000,0000,,just 2X, so this is just\Nthree times 2X, which is 6X,
Dialogue: 0,0:21:05.44,0:21:07.36,Default,,0000,0000,0000,,and there's a Y squared.
Dialogue: 0,0:21:08.75,0:21:12.54,Default,,0000,0000,0000,,And finally, there's just the Y\Ncubed at the end. One Y cubed is
Dialogue: 0,0:21:12.54,0:21:16.96,Default,,0000,0000,0000,,just Y cubed. So there we've\Nexpanded the binomial expression
Dialogue: 0,0:21:16.96,0:21:22.56,Default,,0000,0000,0000,,2X plus Y to the power three in\Njust a couple of lines using
Dialogue: 0,0:21:22.56,0:21:26.100,Default,,0000,0000,0000,,pascals triangle. Let's look at\Nanother one. Suppose this time
Dialogue: 0,0:21:26.100,0:21:31.16,Default,,0000,0000,0000,,we want one plus P different\Nletter just for a change one
Dialogue: 0,0:21:31.16,0:21:33.94,Default,,0000,0000,0000,,plus P or raised to the power 4.
Dialogue: 0,0:21:34.49,0:21:38.94,Default,,0000,0000,0000,,In lots of ways, this is going\Nto be a bit simpler.
Dialogue: 0,0:21:39.63,0:21:43.11,Default,,0000,0000,0000,,Because as we move through the\Nterms from left to right, we
Dialogue: 0,0:21:43.11,0:21:44.85,Default,,0000,0000,0000,,want powers of the first term,
Dialogue: 0,0:21:44.85,0:21:49.20,Default,,0000,0000,0000,,which is one. It won't want to\Nthe Power 4 one to the Power 3,
Dialogue: 0,0:21:49.20,0:21:52.89,Default,,0000,0000,0000,,one to the power two and so on,\Nbut want to any power is still
Dialogue: 0,0:21:52.89,0:21:54.37,Default,,0000,0000,0000,,one that's going to make life
Dialogue: 0,0:21:54.37,0:21:58.14,Default,,0000,0000,0000,,easier for ourselves. So 1 to\Nthe power four is just one.
Dialogue: 0,0:21:58.88,0:22:00.06,Default,,0000,0000,0000,,And then thereafter they'll be
Dialogue: 0,0:22:00.06,0:22:02.59,Default,,0000,0000,0000,,just one. All the way through.
Dialogue: 0,0:22:04.27,0:22:07.68,Default,,0000,0000,0000,,We want the powers of P to\Nincrease. We don't want any
Dialogue: 0,0:22:07.68,0:22:09.10,Default,,0000,0000,0000,,peace in the first term.
Dialogue: 0,0:22:09.87,0:22:11.46,Default,,0000,0000,0000,,We want to be there.
Dialogue: 0,0:22:12.17,0:22:17.15,Default,,0000,0000,0000,,P squared there the next time\Nwill have a P cubed in and the
Dialogue: 0,0:22:17.15,0:22:21.07,Default,,0000,0000,0000,,last term will have a Peter. The\Nfour in these ones.
Dialogue: 0,0:22:21.73,0:22:25.63,Default,,0000,0000,0000,,Are the powers of the first term\None, so 1 to the 4th, one to
Dialogue: 0,0:22:25.63,0:22:29.01,Default,,0000,0000,0000,,three, 1 to the two, 1 to the\None which is just one?
Dialogue: 0,0:22:29.86,0:22:31.35,Default,,0000,0000,0000,,And no ones there at all.
Dialogue: 0,0:22:33.45,0:22:36.29,Default,,0000,0000,0000,,And finally, we want some\Ncoefficients and the
Dialogue: 0,0:22:36.29,0:22:39.48,Default,,0000,0000,0000,,coefficients come from pascals\Ntriangle. This time the row
Dialogue: 0,0:22:39.48,0:22:42.32,Default,,0000,0000,0000,,beginning 1, four. Because of\Nthis powerful here.
Dialogue: 0,0:22:43.06,0:22:48.71,Default,,0000,0000,0000,,So the numbers we\Nwant our 14641.
Dialogue: 0,0:22:49.80,0:22:51.12,Default,,0000,0000,0000,,1.
Dialogue: 0,0:22:52.04,0:22:56.32,Default,,0000,0000,0000,,4.\N6.
Dialogue: 0,0:22:57.45,0:23:02.96,Default,,0000,0000,0000,,4. One, let's\Njust tidy it
Dialogue: 0,0:23:02.96,0:23:05.05,Default,,0000,0000,0000,,up as one.
Dialogue: 0,0:23:06.40,0:23:09.86,Default,,0000,0000,0000,,4 * 1 is just four P.
Dialogue: 0,0:23:09.94,0:23:13.92,Default,,0000,0000,0000,,6 * 1 is 66
Dialogue: 0,0:23:13.92,0:23:20.94,Default,,0000,0000,0000,,P squared. 4 * 1\Nis 4 P cubed.
Dialogue: 0,0:23:20.94,0:23:25.52,Default,,0000,0000,0000,,And last of all, one times Peter\Nthe four is just Peter the four.
Dialogue: 0,0:23:26.29,0:23:30.58,Default,,0000,0000,0000,,Again, another example of a\Nbinomial expression raised to a
Dialogue: 0,0:23:30.58,0:23:35.30,Default,,0000,0000,0000,,power, and we can almost write\Nthe answer straight down using
Dialogue: 0,0:23:35.30,0:23:38.73,Default,,0000,0000,0000,,the triangle instead of\Nmultiplying those brackets out
Dialogue: 0,0:23:38.73,0:23:40.45,Default,,0000,0000,0000,,over and over again.
Dialogue: 0,0:23:40.46,0:23:46.93,Default,,0000,0000,0000,,Now, sometimes either or both of\Nthe terms in the binomial
Dialogue: 0,0:23:46.93,0:23:49.28,Default,,0000,0000,0000,,expression might be negative.
Dialogue: 0,0:23:49.79,0:23:54.20,Default,,0000,0000,0000,,So let's have a look at an\Nexample where one of the terms
Dialogue: 0,0:23:54.20,0:23:56.23,Default,,0000,0000,0000,,is negative. So suppose we want
Dialogue: 0,0:23:56.23,0:23:56.91,Default,,0000,0000,0000,,to expand.
Dialogue: 0,0:23:57.71,0:24:04.03,Default,,0000,0000,0000,,3A.\NMinus 2B, so I've got a term
Dialogue: 0,0:24:04.03,0:24:07.40,Default,,0000,0000,0000,,that's negative now, minus 2B,\Nand let's suppose we want this
Dialogue: 0,0:24:07.40,0:24:08.62,Default,,0000,0000,0000,,to the power 5.
Dialogue: 0,0:24:09.61,0:24:15.01,Default,,0000,0000,0000,,3A minus 2B all raised to\Nthe power 5.
Dialogue: 0,0:24:15.62,0:24:17.90,Default,,0000,0000,0000,,This is going to be a bit more\Ncomplicated this time, so let's
Dialogue: 0,0:24:17.90,0:24:19.12,Default,,0000,0000,0000,,see how we get on with it.
Dialogue: 0,0:24:19.76,0:24:25.12,Default,,0000,0000,0000,,As before. We\Nwant to take our first term.
Dialogue: 0,0:24:25.66,0:24:28.72,Default,,0000,0000,0000,,And raise it to the\Nhighest power, the highest
Dialogue: 0,0:24:28.72,0:24:29.74,Default,,0000,0000,0000,,power being 5.
Dialogue: 0,0:24:30.94,0:24:34.31,Default,,0000,0000,0000,,So our first term will be 3A.
Dialogue: 0,0:24:34.32,0:24:36.35,Default,,0000,0000,0000,,All raised to the power 5.
Dialogue: 0,0:24:36.87,0:24:40.70,Default,,0000,0000,0000,,The next term will have a 3A
Dialogue: 0,0:24:40.70,0:24:45.12,Default,,0000,0000,0000,,in it. And this time it will be\Nraised to the power 4.
Dialogue: 0,0:24:48.91,0:24:53.100,Default,,0000,0000,0000,,There be another term with a 3A\Nin. It'll be 3A to the power 3.
Dialogue: 0,0:24:55.22,0:24:58.25,Default,,0000,0000,0000,,Then 3A to the power 2.
Dialogue: 0,0:24:59.92,0:25:04.29,Default,,0000,0000,0000,,Then 3A to the power one, and\Nthen they'll be a final term
Dialogue: 0,0:25:04.29,0:25:06.64,Default,,0000,0000,0000,,that doesn't have 3A in it at
Dialogue: 0,0:25:06.64,0:25:10.59,Default,,0000,0000,0000,,all. That deals with this\Nfirst term.
Dialogue: 0,0:25:12.81,0:25:14.90,Default,,0000,0000,0000,,Let's deal with\Nthe minus 2B now.
Dialogue: 0,0:25:16.83,0:25:21.09,Default,,0000,0000,0000,,In the first term here, there\Nwon't be any minus two BS at
Dialogue: 0,0:25:21.09,0:25:24.37,Default,,0000,0000,0000,,all, but there after the\Npowers of this term will
Dialogue: 0,0:25:24.37,0:25:28.31,Default,,0000,0000,0000,,increase as we move from left\Nto right exactly as before. So
Dialogue: 0,0:25:28.31,0:25:32.57,Default,,0000,0000,0000,,when we get to the second term\Nhere will need a minus two
Dialogue: 0,0:25:32.57,0:25:32.90,Default,,0000,0000,0000,,fee.
Dialogue: 0,0:25:35.64,0:25:39.98,Default,,0000,0000,0000,,When we get to the next term\Nwill leave minus 2B and we're
Dialogue: 0,0:25:39.98,0:25:41.32,Default,,0000,0000,0000,,going to square it.
Dialogue: 0,0:25:43.18,0:25:47.11,Default,,0000,0000,0000,,Minus 2B raised to the power 3.
Dialogue: 0,0:25:48.63,0:25:51.91,Default,,0000,0000,0000,,Minus two be raised to the power
Dialogue: 0,0:25:51.91,0:25:58.30,Default,,0000,0000,0000,,4. And the last term will be\Nminus two be raised to the power
Dialogue: 0,0:25:58.30,0:26:01.78,Default,,0000,0000,0000,,5. The power five\Ncorresponding to the highest
Dialogue: 0,0:26:01.78,0:26:03.47,Default,,0000,0000,0000,,power in the original\Nproblem.
Dialogue: 0,0:26:06.57,0:26:11.37,Default,,0000,0000,0000,,We also need our coefficients.\NThe numbers in front of each of
Dialogue: 0,0:26:11.37,0:26:12.57,Default,,0000,0000,0000,,these six terms.
Dialogue: 0,0:26:13.42,0:26:18.11,Default,,0000,0000,0000,,The coefficients come from the\Nrow beginning 15.
Dialogue: 0,0:26:18.64,0:26:21.12,Default,,0000,0000,0000,,Because the problem has a\Npower five in it.
Dialogue: 0,0:26:22.33,0:26:26.59,Default,,0000,0000,0000,,The coefficients\Nare one 510-1051.
Dialogue: 0,0:26:27.88,0:26:32.34,Default,,0000,0000,0000,,One 510-1051 so we\Nwant one of those.
Dialogue: 0,0:26:33.81,0:26:36.04,Default,,0000,0000,0000,,Five of those.
Dialogue: 0,0:26:37.18,0:26:40.05,Default,,0000,0000,0000,,Ten of
Dialogue: 0,0:26:40.05,0:26:43.52,Default,,0000,0000,0000,,those. Ten of
Dialogue: 0,0:26:43.52,0:26:47.37,Default,,0000,0000,0000,,those. Five of those, and\Nfinally one of those you can see
Dialogue: 0,0:26:47.37,0:26:50.96,Default,,0000,0000,0000,,now why I left a lot of space\Nwhen I was writing all this
Dialogue: 0,0:26:50.96,0:26:53.79,Default,,0000,0000,0000,,down. There's a lot of things to\Ntidy up in here.
Dialogue: 0,0:26:55.06,0:26:58.85,Default,,0000,0000,0000,,Just to tidy all this up, we\Nneed to remember that when
Dialogue: 0,0:26:58.85,0:27:02.33,Default,,0000,0000,0000,,we raise a negative number\Nto say the power two, the
Dialogue: 0,0:27:02.33,0:27:06.12,Default,,0000,0000,0000,,results going to be positive\Nwhen we raise it to an even
Dialogue: 0,0:27:06.12,0:27:09.60,Default,,0000,0000,0000,,even power, the result would\Nbe positive. So this term is
Dialogue: 0,0:27:09.60,0:27:13.39,Default,,0000,0000,0000,,going to be positive and the\Nminus 2B to the power four
Dialogue: 0,0:27:13.39,0:27:14.65,Default,,0000,0000,0000,,will also become positive.
Dialogue: 0,0:27:15.82,0:27:19.82,Default,,0000,0000,0000,,When we raise it to an odd power\Nlike 3 or the five, the result
Dialogue: 0,0:27:19.82,0:27:23.30,Default,,0000,0000,0000,,is going to be negative. So our\Nanswer is going to have some
Dialogue: 0,0:27:23.30,0:27:24.36,Default,,0000,0000,0000,,positive and some negative
Dialogue: 0,0:27:24.36,0:27:27.62,Default,,0000,0000,0000,,numbers in it. Let's tidy\Nit all up.
Dialogue: 0,0:27:28.72,0:27:31.03,Default,,0000,0000,0000,,Go to Calculator for this,\N'cause I'm going to raise some
Dialogue: 0,0:27:31.03,0:27:32.29,Default,,0000,0000,0000,,of these numbers to some powers.
Dialogue: 0,0:27:33.05,0:27:36.10,Default,,0000,0000,0000,,First of all I want to raise 3\Nto the power 5.
Dialogue: 0,0:27:38.48,0:27:41.92,Default,,0000,0000,0000,,3 to the power five is
Dialogue: 0,0:27:41.92,0:27:45.09,Default,,0000,0000,0000,,243. So I have 243.
Dialogue: 0,0:27:45.93,0:27:48.64,Default,,0000,0000,0000,,A to the power 5.
Dialogue: 0,0:27:48.64,0:27:51.25,Default,,0000,0000,0000,,And it's all multiplied by\None which isn't going to
Dialogue: 0,0:27:51.25,0:27:51.77,Default,,0000,0000,0000,,change anything.
Dialogue: 0,0:27:53.22,0:27:56.16,Default,,0000,0000,0000,,Now here we've got a negative\Nnumber because this is minus 2
Dialogue: 0,0:27:56.16,0:27:59.34,Default,,0000,0000,0000,,be raised to the power one is\Ngoing to be negative, so this
Dialogue: 0,0:27:59.34,0:28:01.06,Default,,0000,0000,0000,,term is going to have a minus
Dialogue: 0,0:28:01.06,0:28:05.47,Default,,0000,0000,0000,,sign at the front. We've got 3\Nto the power 4.
Dialogue: 0,0:28:06.67,0:28:11.74,Default,,0000,0000,0000,,Well, I know 3 squared is 9 and\N9, nine 481, so 3 to the power
Dialogue: 0,0:28:11.74,0:28:14.69,Default,,0000,0000,0000,,four is 81. Five 210
Dialogue: 0,0:28:15.43,0:28:21.02,Default,,0000,0000,0000,,So I'm going to multiply 81 by\N10, which is 810th.
Dialogue: 0,0:28:21.97,0:28:25.65,Default,,0000,0000,0000,,There will be 8 to the power
Dialogue: 0,0:28:25.65,0:28:29.23,Default,,0000,0000,0000,,4. And a single be.
Dialogue: 0,0:28:29.28,0:28:30.56,Default,,0000,0000,0000,,So that's my next term.
Dialogue: 0,0:28:31.28,0:28:35.72,Default,,0000,0000,0000,,Now what have we got left?\NThere's 3 to the power three
Dialogue: 0,0:28:35.72,0:28:38.31,Default,,0000,0000,0000,,which is 3 cubed, which is 27.
Dialogue: 0,0:28:39.23,0:28:42.31,Default,,0000,0000,0000,,Multiplied by two\Nsquared, which is 4.
Dialogue: 0,0:28:45.37,0:28:46.91,Default,,0000,0000,0000,,All multiplied by 10.
Dialogue: 0,0:28:47.95,0:28:50.93,Default,,0000,0000,0000,,Which is 1080.
Dialogue: 0,0:28:50.93,0:28:54.35,Default,,0000,0000,0000,,8 to the
Dialogue: 0,0:28:54.35,0:28:59.76,Default,,0000,0000,0000,,power 3. B to\Nthe power 2.
Dialogue: 0,0:29:01.44,0:29:04.13,Default,,0000,0000,0000,,And here we have two cubed which
Dialogue: 0,0:29:04.13,0:29:07.32,Default,,0000,0000,0000,,is 8. 3 squared which is 9.
Dialogue: 0,0:29:08.30,0:29:14.50,Default,,0000,0000,0000,,9 eight 472 *\N10 is 720.
Dialogue: 0,0:29:14.50,0:29:18.38,Default,,0000,0000,0000,,There will be an A squared from
Dialogue: 0,0:29:18.38,0:29:23.98,Default,,0000,0000,0000,,this term. And not be a B cubed\Nfrom the last time.
Dialogue: 0,0:29:25.26,0:29:27.11,Default,,0000,0000,0000,,What about here?
Dialogue: 0,0:29:27.99,0:29:29.82,Default,,0000,0000,0000,,Well, we've 2 to the power 4.
Dialogue: 0,0:29:30.49,0:29:31.63,Default,,0000,0000,0000,,Which is 16.
Dialogue: 0,0:29:32.20,0:29:34.65,Default,,0000,0000,0000,,5 three is a 15 here.
Dialogue: 0,0:29:35.33,0:29:42.14,Default,,0000,0000,0000,,And 15 * 16 is 240. It'll\Nbe positive because here we been
Dialogue: 0,0:29:42.14,0:29:48.95,Default,,0000,0000,0000,,negative number to an even power\N248 to the power one or just
Dialogue: 0,0:29:48.95,0:29:52.75,Default,,0000,0000,0000,,a. B to the power 4.
Dialogue: 0,0:29:55.04,0:29:59.43,Default,,0000,0000,0000,,And finally. There will be one\Nmore term and that will be minus
Dialogue: 0,0:29:59.43,0:30:01.50,Default,,0000,0000,0000,,2 to the Power 5, which is going
Dialogue: 0,0:30:01.50,0:30:07.28,Default,,0000,0000,0000,,to be negative. 32 B to\Nthe power 5.
Dialogue: 0,0:30:07.40,0:30:11.67,Default,,0000,0000,0000,,And that's the expansion of this\Nrather complicated expression,
Dialogue: 0,0:30:11.67,0:30:16.41,Default,,0000,0000,0000,,which had both positive and\Nnegative quantities in it. And
Dialogue: 0,0:30:16.41,0:30:20.20,Default,,0000,0000,0000,,again, we've used pascals\Ntriangle to do that.
Dialogue: 0,0:30:21.26,0:30:27.30,Default,,0000,0000,0000,,We can use exactly the same\Nmethod even if there are
Dialogue: 0,0:30:27.30,0:30:33.34,Default,,0000,0000,0000,,fractions involved, so let's\Nhave a look at an example where
Dialogue: 0,0:30:33.34,0:30:37.73,Default,,0000,0000,0000,,there's some fractions. Suppose\Nwe want to expand.
Dialogue: 0,0:30:37.73,0:30:44.11,Default,,0000,0000,0000,,This time 1 + 2 over X, so\NI've deliberately put a fraction
Dialogue: 0,0:30:44.11,0:30:47.55,Default,,0000,0000,0000,,in there all to the power 3.
Dialogue: 0,0:30:48.79,0:30:50.21,Default,,0000,0000,0000,,Let's see what happens.
Dialogue: 0,0:30:51.14,0:30:57.60,Default,,0000,0000,0000,,1 + 2 over\NX to the power
Dialogue: 0,0:30:57.60,0:31:03.33,Default,,0000,0000,0000,,3. Well. We start\Nwith one raised to the highest
Dialogue: 0,0:31:03.33,0:31:05.72,Default,,0000,0000,0000,,power which is 1 to the power 3.
Dialogue: 0,0:31:06.46,0:31:07.89,Default,,0000,0000,0000,,Which is still 1.
Dialogue: 0,0:31:08.75,0:31:12.23,Default,,0000,0000,0000,,And once at, any power will\Nstill be one's remove all the
Dialogue: 0,0:31:12.23,0:31:13.39,Default,,0000,0000,0000,,way through the calculation.
Dialogue: 0,0:31:14.87,0:31:21.38,Default,,0000,0000,0000,,Will have two over X raised\Nfirst of all to the power one.
Dialogue: 0,0:31:21.48,0:31:27.20,Default,,0000,0000,0000,,Two over X to\Nthe power 2.
Dialogue: 0,0:31:27.75,0:31:32.47,Default,,0000,0000,0000,,And two over X to the power\Nthree and we stop there. When we
Dialogue: 0,0:31:32.47,0:31:33.82,Default,,0000,0000,0000,,reached the highest power.
Dialogue: 0,0:31:34.47,0:31:36.93,Default,,0000,0000,0000,,Which corresponds to the power\Nin the original problem.
Dialogue: 0,0:31:38.98,0:31:43.53,Default,,0000,0000,0000,,We need the coefficients of each\Nof these terms from pascals
Dialogue: 0,0:31:43.53,0:31:47.26,Default,,0000,0000,0000,,triangle and the row in the\Ntriangle beginning 13.
Dialogue: 0,0:31:48.09,0:31:54.68,Default,,0000,0000,0000,,Those numbers are 1331, so\Nthere's one of these three of
Dialogue: 0,0:31:54.68,0:31:56.60,Default,,0000,0000,0000,,those. Three of those.
Dialogue: 0,0:31:57.22,0:31:59.04,Default,,0000,0000,0000,,I'm one of those.
Dialogue: 0,0:31:59.80,0:32:03.13,Default,,0000,0000,0000,,And all we need to do now is\Ntidy at what we've got.
Dialogue: 0,0:32:04.18,0:32:05.49,Default,,0000,0000,0000,,So there's once.
Dialogue: 0,0:32:07.03,0:32:10.98,Default,,0000,0000,0000,,Two over X to the power one\Nis just two over X. We're
Dialogue: 0,0:32:10.98,0:32:14.63,Default,,0000,0000,0000,,going to multiply it by\Nthree, so 3 twos are six will
Dialogue: 0,0:32:14.63,0:32:16.15,Default,,0000,0000,0000,,have 6 divided by X.
Dialogue: 0,0:32:17.98,0:32:23.33,Default,,0000,0000,0000,,Here there's a 2 squared, which\Nis 4. Multiply it by three so we
Dialogue: 0,0:32:23.33,0:32:28.29,Default,,0000,0000,0000,,have 12 divided by X to the\Npower 2 divided by X squared.
Dialogue: 0,0:32:29.49,0:32:31.57,Default,,0000,0000,0000,,And finally, there's 2 to the
Dialogue: 0,0:32:31.57,0:32:33.93,Default,,0000,0000,0000,,power 3. Which is 8.
Dialogue: 0,0:32:34.84,0:32:40.86,Default,,0000,0000,0000,,And this time it's divided by X\Nto the power 34X cubed.
Dialogue: 0,0:32:41.47,0:32:45.12,Default,,0000,0000,0000,,So that's a simple example which\Nillustrates how we can apply
Dialogue: 0,0:32:45.12,0:32:48.44,Default,,0000,0000,0000,,exactly the same technique even\Nwhen the refraction is involved.
Dialogue: 0,0:32:49.33,0:32:57.26,Default,,0000,0000,0000,,Now, that's not quite\Nthe end of the
Dialogue: 0,0:32:57.26,0:33:02.78,Default,,0000,0000,0000,,story. The problem is, supposing\NI were to ask you to expand a
Dialogue: 0,0:33:02.78,0:33:05.84,Default,,0000,0000,0000,,binomial expression to a very\Nlarge power, suppose I wanted
Dialogue: 0,0:33:05.84,0:33:10.74,Default,,0000,0000,0000,,one plus X to the power 32 or\None plus X to the power 127. You
Dialogue: 0,0:33:10.74,0:33:14.41,Default,,0000,0000,0000,,have an awful lot of rows of\Npascals triangle to generate if
Dialogue: 0,0:33:14.41,0:33:16.55,Default,,0000,0000,0000,,you wanted to do it this way.
Dialogue: 0,0:33:17.25,0:33:20.45,Default,,0000,0000,0000,,Fortunately, there's an\Nalternative way, and it involves
Dialogue: 0,0:33:20.45,0:33:22.45,Default,,0000,0000,0000,,a theorem called the binomial
Dialogue: 0,0:33:22.45,0:33:26.66,Default,,0000,0000,0000,,theorem. So let's just have\Na look at what the binomial
Dialogue: 0,0:33:26.66,0:33:27.32,Default,,0000,0000,0000,,theorem says.
Dialogue: 0,0:33:28.54,0:33:36.00,Default,,0000,0000,0000,,The binomial theorem allows us\Nto develop an expansion of
Dialogue: 0,0:33:36.00,0:33:42.71,Default,,0000,0000,0000,,the binomial expression A+B\Nraised to the power N.
Dialogue: 0,0:33:44.22,0:33:48.51,Default,,0000,0000,0000,,And it allows us to get an\Nexpansion in terms of
Dialogue: 0,0:33:48.51,0:33:52.02,Default,,0000,0000,0000,,decreasing powers of a,\Nexactly as we've seen before.
Dialogue: 0,0:33:53.31,0:33:56.79,Default,,0000,0000,0000,,And increasing powers of B\Nexactly as we've seen before.
Dialogue: 0,0:33:56.79,0:34:02.01,Default,,0000,0000,0000,,And it I'm going to quote the\Ntheorem for the case when N is a
Dialogue: 0,0:34:02.01,0:34:03.05,Default,,0000,0000,0000,,positive whole number.
Dialogue: 0,0:34:04.27,0:34:08.81,Default,,0000,0000,0000,,This theorem will actually work\Nwhen is negative and when it's a
Dialogue: 0,0:34:08.81,0:34:11.45,Default,,0000,0000,0000,,fraction, but only under\Nexceptional circumstances, which
Dialogue: 0,0:34:11.45,0:34:16.37,Default,,0000,0000,0000,,we're not going to discuss here.\NSo in all these examples, N will
Dialogue: 0,0:34:16.37,0:34:18.26,Default,,0000,0000,0000,,be a positive whole number.
Dialogue: 0,0:34:19.43,0:34:25.74,Default,,0000,0000,0000,,Now what\Nthe theorem
Dialogue: 0,0:34:25.74,0:34:28.90,Default,,0000,0000,0000,,says is
Dialogue: 0,0:34:28.90,0:34:34.69,Default,,0000,0000,0000,,this. A+B to the power N is\Ngiven by the following expansion
Dialogue: 0,0:34:34.69,0:34:36.61,Default,,0000,0000,0000,,A to the power N.
Dialogue: 0,0:34:37.23,0:34:40.04,Default,,0000,0000,0000,,Now that looks familiar, doesn't\Nit? Because as in all the
Dialogue: 0,0:34:40.04,0:34:42.58,Default,,0000,0000,0000,,examples we've seen before,\Nwe've taken the first term and
Dialogue: 0,0:34:42.58,0:34:45.39,Default,,0000,0000,0000,,raised it to the highest power.\NThe power in the original
Dialogue: 0,0:34:45.39,0:34:46.92,Default,,0000,0000,0000,,question 8 to the power N.
Dialogue: 0,0:34:47.67,0:34:51.73,Default,,0000,0000,0000,,Then there's a next term, and\Nthe next term will have an A to
Dialogue: 0,0:34:51.73,0:34:53.18,Default,,0000,0000,0000,,the power N minus one.
Dialogue: 0,0:34:53.92,0:34:57.89,Default,,0000,0000,0000,,And a B in it. That's exactly as\Nwe've seen before, because we're
Dialogue: 0,0:34:57.89,0:35:00.73,Default,,0000,0000,0000,,starting to see the terms\Ninvolving be appear and the
Dialogue: 0,0:35:00.73,0:35:02.72,Default,,0000,0000,0000,,powers event at the powers of a
Dialogue: 0,0:35:02.72,0:35:07.02,Default,,0000,0000,0000,,a decreasing. We want a\Ncoefficient in here and the
Dialogue: 0,0:35:07.02,0:35:10.27,Default,,0000,0000,0000,,binomial theorem tells us that\Nthe coefficient is NTH.
Dialogue: 0,0:35:12.14,0:35:15.74,Default,,0000,0000,0000,,The next term.
Dialogue: 0,0:35:15.74,0:35:19.34,Default,,0000,0000,0000,,As an A to the power\NN minus two in it.
Dialogue: 0,0:35:20.54,0:35:23.87,Default,,0000,0000,0000,,Along with the line we had\Nbefore of decreasing the powers
Dialogue: 0,0:35:23.87,0:35:27.51,Default,,0000,0000,0000,,and increasing the power of be\Nwill give us a B squared.
Dialogue: 0,0:35:28.13,0:35:31.85,Default,,0000,0000,0000,,And the binomial theorem\Ntells us the coefficient to
Dialogue: 0,0:35:31.85,0:35:35.98,Default,,0000,0000,0000,,right in here and the\Ncoefficient this time is NN
Dialogue: 0,0:35:35.98,0:35:38.04,Default,,0000,0000,0000,,minus one over 2 factorial.
Dialogue: 0,0:35:39.35,0:35:43.86,Default,,0000,0000,0000,,In case you don't know what this\Nnotation means, 2 factorial
Dialogue: 0,0:35:43.86,0:35:45.50,Default,,0000,0000,0000,,means 2 * 1.
Dialogue: 0,0:35:46.43,0:35:49.44,Default,,0000,0000,0000,,That's called 2 factorial.
Dialogue: 0,0:35:50.37,0:35:57.72,Default,,0000,0000,0000,,And this series goes on and on\Nand on. The next term will be
Dialogue: 0,0:35:57.72,0:36:04.02,Default,,0000,0000,0000,,NN minus one and minus two over\N3 factorial, and there's a
Dialogue: 0,0:36:04.02,0:36:09.80,Default,,0000,0000,0000,,pattern developing here. You\Nsee, here we had an N&NN minus
Dialogue: 0,0:36:09.80,0:36:13.05,Default,,0000,0000,0000,,one. And minus one and minus 2.
Dialogue: 0,0:36:13.55,0:36:19.16,Default,,0000,0000,0000,,With a 3 factorial at the bottom\Nwhere we had a two factor at the
Dialogue: 0,0:36:19.16,0:36:22.90,Default,,0000,0000,0000,,bottom before 3 factorial means\N3 * 2 * 1.
Dialogue: 0,0:36:23.87,0:36:28.88,Default,,0000,0000,0000,,The power of a will be 1 less\Nagain, which this time will be A
Dialogue: 0,0:36:28.88,0:36:30.55,Default,,0000,0000,0000,,to the N minus three.
Dialogue: 0,0:36:31.47,0:36:35.26,Default,,0000,0000,0000,,And we want to power of bee\Nwhich is B to the power 3.
Dialogue: 0,0:36:36.92,0:36:40.50,Default,,0000,0000,0000,,So all the way through this\Ntheorem you'll see the powers of
Dialogue: 0,0:36:40.50,0:36:45.47,Default,,0000,0000,0000,,a are decreasing. And the powers\Nof B are increasing. Now this
Dialogue: 0,0:36:45.47,0:36:50.99,Default,,0000,0000,0000,,series goes on and on and on\Nuntil we reach the term B to the
Dialogue: 0,0:36:50.99,0:36:55.78,Default,,0000,0000,0000,,power N. When it stops. So this\Nis a finite series. It stops
Dialogue: 0,0:36:55.78,0:36:57.98,Default,,0000,0000,0000,,after a finite number of terms.
Dialogue: 0,0:36:58.91,0:37:02.33,Default,,0000,0000,0000,,Now, the theorems often quoted\Nin this form, but it's also
Dialogue: 0,0:37:02.33,0:37:03.89,Default,,0000,0000,0000,,often quoted in a slightly
Dialogue: 0,0:37:03.89,0:37:08.35,Default,,0000,0000,0000,,simpler form. And it's quoted in\Nthe form for which a is the
Dialogue: 0,0:37:08.35,0:37:09.78,Default,,0000,0000,0000,,simple value of just one.
Dialogue: 0,0:37:10.41,0:37:16.26,Default,,0000,0000,0000,,And B is X. Now when a is one,\Nall of these A to the power ends
Dialogue: 0,0:37:16.26,0:37:22.11,Default,,0000,0000,0000,,or A to the N minus one A to the\NN minus two. Each one of those
Dialogue: 0,0:37:22.11,0:37:26.23,Default,,0000,0000,0000,,terms will just simplify to the\Nnumber one, so the whole thing
Dialogue: 0,0:37:26.23,0:37:29.67,Default,,0000,0000,0000,,looks simpler. So let's write\Ndown the binomial theorem again
Dialogue: 0,0:37:29.67,0:37:33.46,Default,,0000,0000,0000,,for the special case when a is\None and these X.
Dialogue: 0,0:37:33.50,0:37:40.38,Default,,0000,0000,0000,,This time will get one\Nplus X raised to the
Dialogue: 0,0:37:40.38,0:37:43.39,Default,,0000,0000,0000,,power N. Is equal to.
Dialogue: 0,0:37:44.04,0:37:44.88,Default,,0000,0000,0000,,1.
Dialogue: 0,0:37:45.98,0:37:50.07,Default,,0000,0000,0000,,Plus N.\NX.
Dialogue: 0,0:37:51.87,0:37:56.15,Default,,0000,0000,0000,,Plus NN minus one over 2\Nfactorial X squared, and you can
Dialogue: 0,0:37:56.15,0:37:59.72,Default,,0000,0000,0000,,see what's happening. This\Nsecond term X is starting to
Dialogue: 0,0:37:59.72,0:38:03.65,Default,,0000,0000,0000,,appear and and its powers\Nincreasing as we move from left
Dialogue: 0,0:38:03.65,0:38:08.29,Default,,0000,0000,0000,,to right. So even X&X squared\Nthe next time will have an X
Dialogue: 0,0:38:08.29,0:38:13.29,Default,,0000,0000,0000,,cubed in it, one to any power is\Nstill one, so I don't actually
Dialogue: 0,0:38:13.29,0:38:15.08,Default,,0000,0000,0000,,need to write it down.
Dialogue: 0,0:38:15.71,0:38:22.12,Default,,0000,0000,0000,,The next term will\Nbe NN minus one and
Dialogue: 0,0:38:22.12,0:38:27.10,Default,,0000,0000,0000,,minus two over 3\Nfactorial X cubed.
Dialogue: 0,0:38:28.33,0:38:33.65,Default,,0000,0000,0000,,The next term will be NN minus\None and minus 2 N minus three
Dialogue: 0,0:38:33.65,0:38:39.35,Default,,0000,0000,0000,,over 4 factorial X to the four,\Nand this will go on and on until
Dialogue: 0,0:38:39.35,0:38:43.91,Default,,0000,0000,0000,,eventually you'll get to the\Nstage where you get to the last
Dialogue: 0,0:38:43.91,0:38:49.23,Default,,0000,0000,0000,,term raised to the highest power\Nyou'll get to X to the power N,
Dialogue: 0,0:38:49.23,0:38:51.13,Default,,0000,0000,0000,,and the series will stop.
Dialogue: 0,0:38:52.12,0:38:55.29,Default,,0000,0000,0000,,So this is a slightly simpler\Nform of the theorem, and it's
Dialogue: 0,0:38:55.29,0:38:56.61,Default,,0000,0000,0000,,often quoted in this form.
Dialogue: 0,0:38:57.26,0:39:02.56,Default,,0000,0000,0000,,Now let's use it to examine some\Nbinomial expressions that you're
Dialogue: 0,0:39:02.56,0:39:04.49,Default,,0000,0000,0000,,already very familiar with.
Dialogue: 0,0:39:04.49,0:39:08.99,Default,,0000,0000,0000,,Let's suppose we want to expand\None plus X or raised to the
Dialogue: 0,0:39:08.99,0:39:13.14,Default,,0000,0000,0000,,power two. Now I've written down\Nthe theorem again so we can
Dialogue: 0,0:39:13.14,0:39:18.33,Default,,0000,0000,0000,,refer to it and this is printed\Nin the notes. If you want to use
Dialogue: 0,0:39:18.33,0:39:20.06,Default,,0000,0000,0000,,the one in the notes.
Dialogue: 0,0:39:21.30,0:39:27.04,Default,,0000,0000,0000,,So we've one plus X to the power\NN. In our problem, we've got one
Dialogue: 0,0:39:27.04,0:39:33.56,Default,,0000,0000,0000,,plus X to the power two, so all\Nwe have to do is let NB two in
Dialogue: 0,0:39:33.56,0:39:35.47,Default,,0000,0000,0000,,all of this formula through
Dialogue: 0,0:39:35.47,0:39:40.46,Default,,0000,0000,0000,,here. So let's see what we get\Nor from the theorem.
Dialogue: 0,0:39:40.98,0:39:44.04,Default,,0000,0000,0000,,The first thing will write down\Nis just the one.
Dialogue: 0,0:39:45.68,0:39:52.90,Default,,0000,0000,0000,,Then we want NX, but N IS\Ntwo, so will just put plus 2X.
Dialogue: 0,0:39:52.91,0:39:56.63,Default,,0000,0000,0000,,And then the next term we want\Nis going to be a term
Dialogue: 0,0:39:56.63,0:39:59.77,Default,,0000,0000,0000,,involving X to the power two,\Nbut that's the highest power
Dialogue: 0,0:39:59.77,0:40:03.49,Default,,0000,0000,0000,,we want because we've got a\Npower to in here. We want to
Dialogue: 0,0:40:03.49,0:40:07.21,Default,,0000,0000,0000,,stop when we get to X to the\Npower two, so we're actually
Dialogue: 0,0:40:07.21,0:40:10.93,Default,,0000,0000,0000,,already at the end with the\Nnext term, and we just want an
Dialogue: 0,0:40:10.93,0:40:13.22,Default,,0000,0000,0000,,X to the power two on its own.
Dialogue: 0,0:40:14.72,0:40:18.39,Default,,0000,0000,0000,,1 + 2 X plus X squared and\Nthat's the expansion that
Dialogue: 0,0:40:18.39,0:40:21.45,Default,,0000,0000,0000,,you're already very familiar\Nwith, and you'll notice in it
Dialogue: 0,0:40:21.45,0:40:25.43,Default,,0000,0000,0000,,that the powers of X increase\Nas we move through from left to
Dialogue: 0,0:40:25.43,0:40:29.10,Default,,0000,0000,0000,,right, and there's powers of\None in there, but we don't see
Dialogue: 0,0:40:29.10,0:40:32.47,Default,,0000,0000,0000,,them, and the one 2 one other\Nnumbers in pascals triangle.
Dialogue: 0,0:40:33.97,0:40:40.26,Default,,0000,0000,0000,,Let's look at the theorem for\Nthe case when is 3, let's expand
Dialogue: 0,0:40:40.26,0:40:43.65,Default,,0000,0000,0000,,one plus X to the power 3.
Dialogue: 0,0:40:44.81,0:40:47.46,Default,,0000,0000,0000,,I'm going to use the theorem\Nagain, but this time we're
Dialogue: 0,0:40:47.46,0:40:48.67,Default,,0000,0000,0000,,going to let NB 3.
Dialogue: 0,0:40:51.02,0:40:54.61,Default,,0000,0000,0000,,So we want 1 + 3
Dialogue: 0,0:40:54.61,0:41:01.22,Default,,0000,0000,0000,,X. And then\Nwe want 3.
Dialogue: 0,0:41:01.82,0:41:07.60,Default,,0000,0000,0000,,3 - 1 three minus\None is 2.
Dialogue: 0,0:41:08.48,0:41:11.25,Default,,0000,0000,0000,,All divided by 2 factorial.
Dialogue: 0,0:41:12.13,0:41:14.76,Default,,0000,0000,0000,,Than an X squared.
Dialogue: 0,0:41:16.15,0:41:19.78,Default,,0000,0000,0000,,And then the next term will be a\Nterm involving X cubed, which is
Dialogue: 0,0:41:19.78,0:41:22.88,Default,,0000,0000,0000,,the term that we stop with\Nbecause we're only working 2X to
Dialogue: 0,0:41:22.88,0:41:26.51,Default,,0000,0000,0000,,the power three here. So the\Nlast term will be just a plus X
Dialogue: 0,0:41:26.51,0:41:31.76,Default,,0000,0000,0000,,cubed. We can tie this up to 1\N+ 3 X.
Dialogue: 0,0:41:32.63,0:41:37.43,Default,,0000,0000,0000,,2 factorial is 2 * 1, which is\Njust two little cancel with the
Dialogue: 0,0:41:37.43,0:41:42.23,Default,,0000,0000,0000,,two at the top, so will be left\Nwith just three X squared, and
Dialogue: 0,0:41:42.23,0:41:43.61,Default,,0000,0000,0000,,finally an X cubed.
Dialogue: 0,0:41:43.65,0:41:47.27,Default,,0000,0000,0000,,And again, that's something that\Nyou're already very familiar
Dialogue: 0,0:41:47.27,0:41:50.48,Default,,0000,0000,0000,,with. You'll notice the\Ncoefficients, the 1331 other
Dialogue: 0,0:41:50.48,0:41:55.31,Default,,0000,0000,0000,,numbers we've seen many times in\Npascals triangle the powers of X
Dialogue: 0,0:41:55.31,0:42:00.62,Default,,0000,0000,0000,,increase. As we move from the\Nleft to the right, and this is a
Dialogue: 0,0:42:00.62,0:42:04.64,Default,,0000,0000,0000,,finite series, it stops when we\Nget to the term involving X
Dialogue: 0,0:42:04.64,0:42:07.32,Default,,0000,0000,0000,,cubed corresponding to this\Nhighest power over there.
Dialogue: 0,0:42:07.60,0:42:14.81,Default,,0000,0000,0000,,Now.\NThat suppose we want to look at
Dialogue: 0,0:42:14.81,0:42:18.62,Default,,0000,0000,0000,,it and more complicated problem.\NSuppose we want to workout one
Dialogue: 0,0:42:18.62,0:42:21.39,Default,,0000,0000,0000,,plus X to the power 32. Now you
Dialogue: 0,0:42:21.39,0:42:25.08,Default,,0000,0000,0000,,would never. Use pascals\Ntriangle to attempt this problem
Dialogue: 0,0:42:25.08,0:42:28.94,Default,,0000,0000,0000,,because you'd have to generate\Nso many rows of the triangle,
Dialogue: 0,0:42:28.94,0:42:31.04,Default,,0000,0000,0000,,but we can use the binomial
Dialogue: 0,0:42:31.04,0:42:35.03,Default,,0000,0000,0000,,theorem. What I'm going to do\Nis I'm going to write down
Dialogue: 0,0:42:35.03,0:42:37.75,Default,,0000,0000,0000,,the first three terms of the\Nseries using the binomial
Dialogue: 0,0:42:37.75,0:42:41.29,Default,,0000,0000,0000,,theorem, and I'm going to use\Nit with N being equal to 32.
Dialogue: 0,0:42:43.29,0:42:50.36,Default,,0000,0000,0000,,So we're putting any 32 in. The\Ntheorem will get 1 + 32 X.
Dialogue: 0,0:42:50.36,0:42:53.39,Default,,0000,0000,0000,,That's the one plus the NX.
Dialogue: 0,0:42:54.40,0:42:57.41,Default,,0000,0000,0000,,We want an which is 32.
Dialogue: 0,0:42:58.39,0:43:01.13,Default,,0000,0000,0000,,And minus one which will be 31.
Dialogue: 0,0:43:01.73,0:43:04.04,Default,,0000,0000,0000,,All over 2 factorial.
Dialogue: 0,0:43:05.38,0:43:06.49,Default,,0000,0000,0000,,X squared
Dialogue: 0,0:43:07.50,0:43:11.92,Default,,0000,0000,0000,,And we know that this series\Nwill go on and on until we
Dialogue: 0,0:43:11.92,0:43:16.00,Default,,0000,0000,0000,,reached the term, the last term\Nbeing X to the power 32.
Dialogue: 0,0:43:16.66,0:43:20.91,Default,,0000,0000,0000,,But I only want to look at\Nthe first three terms here in
Dialogue: 0,0:43:20.91,0:43:24.51,Default,,0000,0000,0000,,this problem, so the first\Nthree terms are just going to
Dialogue: 0,0:43:24.51,0:43:29.41,Default,,0000,0000,0000,,be 1 + 32 X and we want to\Nsimplify this. We've got 32 *
Dialogue: 0,0:43:29.41,0:43:31.38,Default,,0000,0000,0000,,31 and then divided by two.
Dialogue: 0,0:43:34.94,0:43:38.22,Default,,0000,0000,0000,,Which is 496.
Dialogue: 0,0:43:38.31,0:43:43.00,Default,,0000,0000,0000,,And I just put some dots there\Nto show that this series goes on
Dialogue: 0,0:43:43.00,0:43:46.68,Default,,0000,0000,0000,,a lot further than the terms\Nthat I've just written down
Dialogue: 0,0:43:46.68,0:43:53.31,Default,,0000,0000,0000,,there. I'm going to have a look\Nat a couple more examples with
Dialogue: 0,0:43:53.31,0:43:58.29,Default,,0000,0000,0000,,some ingenuity. We can use the\Ntheorem in a slightly different
Dialogue: 0,0:43:58.29,0:44:02.82,Default,,0000,0000,0000,,form. Suppose we want to expand\Nthis binomial expression this
Dialogue: 0,0:44:02.82,0:44:09.16,Default,,0000,0000,0000,,time, I'm going to look at one\Nplus Y divided by 3. All raised
Dialogue: 0,0:44:09.16,0:44:14.14,Default,,0000,0000,0000,,to the power 10 and suppose that\NI'm interested. I'm interested
Dialogue: 0,0:44:14.14,0:44:15.96,Default,,0000,0000,0000,,in generating the first.
Dialogue: 0,0:44:17.00,0:44:23.37,Default,,0000,0000,0000,,Four terms. Let's see how we can\Ndo that. Well, we've got our
Dialogue: 0,0:44:23.37,0:44:28.52,Default,,0000,0000,0000,,theorem. I've written it down\Nagain here for us in terms of
Dialogue: 0,0:44:28.52,0:44:35.38,Default,,0000,0000,0000,,one plus X to the power N. We\Ncan use it in this problem if we
Dialogue: 0,0:44:35.38,0:44:36.67,Default,,0000,0000,0000,,replace every X.
Dialogue: 0,0:44:37.20,0:44:40.09,Default,,0000,0000,0000,,In the theorem with a Y over 3.
Dialogue: 0,0:44:41.06,0:44:45.34,Default,,0000,0000,0000,,So everywhere there's an X in\Nthe theorem, I'm going to write
Dialogue: 0,0:44:45.34,0:44:49.63,Default,,0000,0000,0000,,Y divided by three and then the\Npattern will match exactly what
Dialogue: 0,0:44:49.63,0:44:54.27,Default,,0000,0000,0000,,we have in the theorem ends\Ngoing to be 10 in this problem.
Dialogue: 0,0:44:54.27,0:44:59.66,Default,,0000,0000,0000,,So let's see what we get will\Nhave one plus Y over three
Dialogue: 0,0:44:59.66,0:45:02.57,Default,,0000,0000,0000,,raised to the power 10 is equal
Dialogue: 0,0:45:02.57,0:45:06.20,Default,,0000,0000,0000,,to. Well, we start with a one
Dialogue: 0,0:45:06.20,0:45:10.06,Default,,0000,0000,0000,,as always. Then we
Dialogue: 0,0:45:10.06,0:45:14.04,Default,,0000,0000,0000,,want NX. And
Dialogue: 0,0:45:14.04,0:45:18.73,Default,,0000,0000,0000,,it's 10. And we said that\Ninstead of X, but replacing the
Dialogue: 0,0:45:18.73,0:45:20.45,Default,,0000,0000,0000,,X with a Y over 3.
Dialogue: 0,0:45:20.96,0:45:24.18,Default,,0000,0000,0000,,So we have a Y over three there.
Dialogue: 0,0:45:25.47,0:45:30.39,Default,,0000,0000,0000,,What's the next term we want NN\Nminus one over 2 factorial?
Dialogue: 0,0:45:31.11,0:45:32.52,Default,,0000,0000,0000,,Which is 10.
Dialogue: 0,0:45:34.15,0:45:38.21,Default,,0000,0000,0000,,10 - 1 is 9 over 2\Nfactorial.
Dialogue: 0,0:45:39.61,0:45:45.12,Default,,0000,0000,0000,,And then we'd want an X squared.\NSo in this case we want X being
Dialogue: 0,0:45:45.12,0:45:46.22,Default,,0000,0000,0000,,why over 3?
Dialogue: 0,0:45:46.79,0:45:49.92,Default,,0000,0000,0000,,All square
Dialogue: 0,0:45:51.69,0:45:55.34,Default,,0000,0000,0000,,I want to generate one more term\N'cause I said I want to look for
Dialogue: 0,0:45:55.34,0:45:58.74,Default,,0000,0000,0000,,four terms, so the next term is\Ngoing to be an which was 10.
Dialogue: 0,0:46:00.03,0:46:02.26,Default,,0000,0000,0000,,N minus one which is 9.
Dialogue: 0,0:46:02.80,0:46:06.88,Default,,0000,0000,0000,,And minus two, which is 8 and\Nthis time over 3 factorial.
Dialogue: 0,0:46:07.45,0:46:13.44,Default,,0000,0000,0000,,So I'm here NN minus one and\Nminus two over 3 factorial and
Dialogue: 0,0:46:13.44,0:46:20.36,Default,,0000,0000,0000,,we want X cubed X is Y over\Nthree, so we want why over 3
Dialogue: 0,0:46:20.36,0:46:26.08,Default,,0000,0000,0000,,cubed. And the series goes on\Nand on. Let's just tidy up what
Dialogue: 0,0:46:26.08,0:46:27.83,Default,,0000,0000,0000,,we've got. There's one.
Dialogue: 0,0:46:29.07,0:46:32.66,Default,,0000,0000,0000,,That'll be 10, why over 3?
Dialogue: 0,0:46:32.66,0:46:37.07,Default,,0000,0000,0000,,What if we got in here? Well,\Nthere's a 3 square at the bottom
Dialogue: 0,0:46:37.07,0:46:41.80,Default,,0000,0000,0000,,which is 9, and there's a 9 at\Nthe top, so the three squared in
Dialogue: 0,0:46:41.80,0:46:44.00,Default,,0000,0000,0000,,here is going to cancel with the
Dialogue: 0,0:46:44.00,0:46:46.61,Default,,0000,0000,0000,,nine there. 2 factorial
Dialogue: 0,0:46:47.21,0:46:48.21,Default,,0000,0000,0000,,Is just two.
Dialogue: 0,0:46:49.43,0:46:55.42,Default,,0000,0000,0000,,And choosing to 10 is 5, so will\Nhave five 5 squared.
Dialogue: 0,0:46:56.21,0:47:02.31,Default,,0000,0000,0000,,And then this is a bit more\Ncomplicated. We've got a 3
Dialogue: 0,0:47:02.31,0:47:05.35,Default,,0000,0000,0000,,factorial which is 3 * 2.
Dialogue: 0,0:47:05.36,0:47:10.94,Default,,0000,0000,0000,,And three cubed. At the bottom\Nthere, which is 3 * 3 * 3. Some
Dialogue: 0,0:47:10.94,0:47:16.15,Default,,0000,0000,0000,,of this will cancel down. 3 * 3\Nwill cancel, with the nine in
Dialogue: 0,0:47:16.15,0:47:21.32,Default,,0000,0000,0000,,here. The two will cancel their\Nwith the eight will have four
Dialogue: 0,0:47:21.32,0:47:26.16,Default,,0000,0000,0000,,and let's see what we're left\Nwith at the top will have 10 *
Dialogue: 0,0:47:26.16,0:47:27.54,Default,,0000,0000,0000,,4, which is 40.
Dialogue: 0,0:47:28.14,0:47:34.40,Default,,0000,0000,0000,,And at the bottom will have 3\N* 3, which is 9.
Dialogue: 0,0:47:34.96,0:47:37.89,Default,,0000,0000,0000,,And they'll be a Y cubed.
Dialogue: 0,0:47:37.89,0:47:44.46,Default,,0000,0000,0000,,So altogether we've 1 + 10 Y\Nover 3 five Y squared, 40 over 9
Dialogue: 0,0:47:44.46,0:47:50.15,Default,,0000,0000,0000,,Y cubed, and those are the first\Nfour terms of a series which
Dialogue: 0,0:47:50.15,0:47:54.97,Default,,0000,0000,0000,,will actually continue until you\Nget to a term involving the
Dialogue: 0,0:47:54.97,0:48:00.67,Default,,0000,0000,0000,,highest power, which will be a Y\Nover 3 to the power 10.
Dialogue: 0,0:48:00.90,0:48:04.43,Default,,0000,0000,0000,,So you can still use the theorem\Nin slightly different form if
Dialogue: 0,0:48:04.43,0:48:08.25,Default,,0000,0000,0000,,you use a bit of ingenuity. Want\Nto look at one final example
Dialogue: 0,0:48:08.25,0:48:14.95,Default,,0000,0000,0000,,before we finish? And this time\NI want to look at the example 3
Dialogue: 0,0:48:14.95,0:48:16.39,Default,,0000,0000,0000,,- 5 Z.
Dialogue: 0,0:48:16.39,0:48:20.24,Default,,0000,0000,0000,,To the power 40 again, it's an\Nexample where you wouldn't want
Dialogue: 0,0:48:20.24,0:48:24.09,Default,,0000,0000,0000,,to use pascals triangle because\Nthe power for teens too high and
Dialogue: 0,0:48:24.09,0:48:27.30,Default,,0000,0000,0000,,you have too many rose to\Ngenerate in your triangle.
Dialogue: 0,0:48:27.88,0:48:31.29,Default,,0000,0000,0000,,I'm going to use the original\Nform of the theorem, the One I
Dialogue: 0,0:48:31.29,0:48:33.38,Default,,0000,0000,0000,,have here in terms of A+B to the
Dialogue: 0,0:48:33.38,0:48:36.91,Default,,0000,0000,0000,,power N. A will be 3.
Dialogue: 0,0:48:38.63,0:48:42.72,Default,,0000,0000,0000,,Now B is a negative number be\Nwill be minus five said.
Dialogue: 0,0:48:44.77,0:48:46.33,Default,,0000,0000,0000,,Ends going to be 14.
Dialogue: 0,0:48:46.92,0:48:50.27,Default,,0000,0000,0000,,But we can still use the\Ntheorem. Let's see what
Dialogue: 0,0:48:50.27,0:48:53.62,Default,,0000,0000,0000,,happens and in this problem\NI'm just going to generate
Dialogue: 0,0:48:53.62,0:48:54.96,Default,,0000,0000,0000,,the first three terms.
Dialogue: 0,0:48:56.26,0:49:02.51,Default,,0000,0000,0000,,OK so A is 3 and we\Nwant to raise the three.
Dialogue: 0,0:49:02.52,0:49:04.99,Default,,0000,0000,0000,,To the highest power which is
Dialogue: 0,0:49:04.99,0:49:09.78,Default,,0000,0000,0000,,14. So my first term is 3 to\Nthe power 14.
Dialogue: 0,0:49:12.07,0:49:14.64,Default,,0000,0000,0000,,My second term is this one.
Dialogue: 0,0:49:15.40,0:49:16.48,Default,,0000,0000,0000,,Begins with an N.
Dialogue: 0,0:49:17.86,0:49:21.05,Default,,0000,0000,0000,,The power in the original\Nexpression, which was 14.
Dialogue: 0,0:49:21.94,0:49:26.70,Default,,0000,0000,0000,,Multiplied by A\Nto the power N
Dialogue: 0,0:49:26.70,0:49:29.42,Default,,0000,0000,0000,,minus one AS 3.
Dialogue: 0,0:49:30.62,0:49:35.77,Default,,0000,0000,0000,,And we want to raise it to the\Npower N minus 114 - 1 is 30.
Dialogue: 0,0:49:36.43,0:49:37.84,Default,,0000,0000,0000,,And we want to be.
Dialogue: 0,0:49:39.50,0:49:41.68,Default,,0000,0000,0000,,B is minus five set.
Dialogue: 0,0:49:42.32,0:49:47.06,Default,,0000,0000,0000,,So our second term looking ahead\Nis going to be negative because
Dialogue: 0,0:49:47.06,0:49:49.43,Default,,0000,0000,0000,,of that minus five in there.
Dialogue: 0,0:49:50.64,0:49:55.70,Default,,0000,0000,0000,,My third term and I'll stop\Nafter the third term, his N,
Dialogue: 0,0:49:55.70,0:49:56.97,Default,,0000,0000,0000,,which is 14.
Dialogue: 0,0:49:57.60,0:50:03.40,Default,,0000,0000,0000,,And minus one which is 13\Nall over 2 factorial.
Dialogue: 0,0:50:04.70,0:50:10.17,Default,,0000,0000,0000,,A to the power N minus two will\Nbe 3 to the power N minus two
Dialogue: 0,0:50:10.17,0:50:12.91,Default,,0000,0000,0000,,will be 14 - 2 which is 12.
Dialogue: 0,0:50:13.56,0:50:16.56,Default,,0000,0000,0000,,And finally AB, which\Nwas minus five said.
Dialogue: 0,0:50:18.31,0:50:20.24,Default,,0000,0000,0000,,Raised to the power 2.
Dialogue: 0,0:50:20.85,0:50:25.16,Default,,0000,0000,0000,,And we know this goes on and on\Nuntil we reach term instead to
Dialogue: 0,0:50:25.16,0:50:28.55,Default,,0000,0000,0000,,the power 14. But we've only\Nwritten down the first three
Dialogue: 0,0:50:28.55,0:50:32.25,Default,,0000,0000,0000,,terms there. Perhaps we should\Njust tidy it up a little bit.
Dialogue: 0,0:50:32.25,0:50:35.12,Default,,0000,0000,0000,,There's a 3 to the power 14 at
Dialogue: 0,0:50:35.12,0:50:42.58,Default,,0000,0000,0000,,the beginning. There's a minus\Nfive here, minus 5. Four
Dialogue: 0,0:50:42.58,0:50:45.58,Default,,0000,0000,0000,,teens are minus 70.
Dialogue: 0,0:50:45.58,0:50:48.84,Default,,0000,0000,0000,,As a 3 to the power 13, let me\Njust leave it like that for the
Dialogue: 0,0:50:48.84,0:50:51.45,Default,,0000,0000,0000,,time being. And then\Nthey'll be as Ed.
Dialogue: 0,0:50:54.13,0:50:56.82,Default,,0000,0000,0000,,Over here there's a 3 to the
Dialogue: 0,0:50:56.82,0:51:00.31,Default,,0000,0000,0000,,power 12. That's this term in
Dialogue: 0,0:51:00.31,0:51:05.51,Default,,0000,0000,0000,,here. And I'm reaching for my\NCalculator again because this is
Dialogue: 0,0:51:05.51,0:51:08.93,Default,,0000,0000,0000,,a bit more complicated. We have\Ngot a 14.
Dialogue: 0,0:51:09.53,0:51:11.25,Default,,0000,0000,0000,,Multiplied by 13.
Dialogue: 0,0:51:12.45,0:51:15.83,Default,,0000,0000,0000,,When multiplied by 5 squared,\Nwhich is 25.
Dialogue: 0,0:51:17.88,0:51:23.52,Default,,0000,0000,0000,,And divided by the two factorial\Nthat's divided by two, this will
Dialogue: 0,0:51:23.52,0:51:25.87,Default,,0000,0000,0000,,be multiplied by two 275.
Dialogue: 0,0:51:26.45,0:51:31.48,Default,,0000,0000,0000,,And this expression will be\Npositive because we've got a
Dialogue: 0,0:51:31.48,0:51:32.99,Default,,0000,0000,0000,,minus 5 squared.
Dialogue: 0,0:51:34.17,0:51:38.68,Default,,0000,0000,0000,,And we need to remember to\Ninclude zed squared in there.
Dialogue: 0,0:51:39.21,0:51:42.40,Default,,0000,0000,0000,,OK, we observe as before that\Nthe powers of zed are increasing
Dialogue: 0,0:51:42.40,0:51:46.13,Default,,0000,0000,0000,,as we move from the left to the\Nright. Now we could leave it
Dialogue: 0,0:51:46.13,0:51:49.85,Default,,0000,0000,0000,,like that. I'm just going to\Ntidy it up and write it in a
Dialogue: 0,0:51:49.85,0:51:52.78,Default,,0000,0000,0000,,slightly different form because\Nthis is often the way you see
Dialogue: 0,0:51:52.78,0:51:56.23,Default,,0000,0000,0000,,answers in the back of textbooks\Nor people ask you to give an
Dialogue: 0,0:51:56.23,0:51:59.69,Default,,0000,0000,0000,,answer in a particular form and\Nthe form I'm going to write it
Dialogue: 0,0:51:59.69,0:52:03.68,Default,,0000,0000,0000,,in is one obtained by taking out\Na factor of 3 to the power 40.
Dialogue: 0,0:52:04.74,0:52:08.52,Default,,0000,0000,0000,,If I take her three to 14 out\Nfrom the first term, I'll be
Dialogue: 0,0:52:08.52,0:52:09.60,Default,,0000,0000,0000,,just left with one.
Dialogue: 0,0:52:10.61,0:52:13.97,Default,,0000,0000,0000,,Now 3 to the 13. In the second
Dialogue: 0,0:52:13.97,0:52:18.65,Default,,0000,0000,0000,,term. But if I multiply top and\Nbottom by three, I'll have a 3
Dialogue: 0,0:52:18.65,0:52:20.89,Default,,0000,0000,0000,,to the 14th at the top, which I
Dialogue: 0,0:52:20.89,0:52:25.15,Default,,0000,0000,0000,,can take out. But have\Nmultiplied the bottom by
Dialogue: 0,0:52:25.15,0:52:29.55,Default,,0000,0000,0000,,three as well, which will\Nleave me with minus 70 Zedd
Dialogue: 0,0:52:29.55,0:52:30.75,Default,,0000,0000,0000,,divided by three.
Dialogue: 0,0:52:32.88,0:52:34.89,Default,,0000,0000,0000,,Here with a 3 to the power 12.
Dialogue: 0,0:52:35.42,0:52:37.59,Default,,0000,0000,0000,,And I want to take out a 3 to
Dialogue: 0,0:52:37.59,0:52:42.91,Default,,0000,0000,0000,,14. If I multiply the top and\Nbottom by three squared or nine,
Dialogue: 0,0:52:42.91,0:52:45.71,Default,,0000,0000,0000,,I affectively get a 3 to 14 in
Dialogue: 0,0:52:45.71,0:52:50.88,Default,,0000,0000,0000,,this term. So I'm multiplying\Ntop and bottom by 9, taking
Dialogue: 0,0:52:50.88,0:52:56.23,Default,,0000,0000,0000,,the three to the 14 out, and\Nthat will leave me here with
Dialogue: 0,0:52:56.23,0:52:57.88,Default,,0000,0000,0000,,two 275 over 9.
Dialogue: 0,0:52:58.98,0:53:04.01,Default,,0000,0000,0000,,Said squad And this series\Ncontinues. As I said before,
Dialogue: 0,0:53:04.01,0:53:08.64,Default,,0000,0000,0000,,until you get to a term\Ninvolving zed to the power 14,
Dialogue: 0,0:53:08.64,0:53:12.50,Default,,0000,0000,0000,,but those are the first three\Nterms of the series.