We're on problem 67.
They ask us how many terms of the binomial expansion of x
squared plus 2y to the third to the
twentieth power contain?
And so right when you see, oh my god, to the twentieth
power, that will take me forever to do, and then I'll
have to count the terms. But remember, they're just asking
you how many terms. They're not asking
you to find the expansion.
So you just have to say, oh boy, let me just think about
this a little bit.
And think of it this way: x plus y to the first power has
how many terms?
It has-- well it's just x plus y.
And so it has 2 terms. x plus y, squared.
That is equal to x squared plus 2xy plus y squared.
So that has 3 terms. If I were to do x plus e The third
power, I could do a binomial expansion, I could do it with
Pascal's Triangle.
Let me do it with Pascal's Triangle.
I have the 1 and the 1.
That's to the first power.
The second power, those are the coefficients.
The third power, the coefficients become 1, 3--
because 1 plus 2 is 3-- 3, 1.
And actually I don't even have to expand it out.
Ill do it just for a little practice.
It x to the third plus 3x squared y plus 3xy squared
plus y to the third.
I just used these.
But when you take any binomial and you take it to the third
power, you end up with 4 terms. And depending on how
you think of it, if you think of it with Pascal's Triangle,
every time we go down we're just going to be adding
another term.
When we take the fourth power it's going to have 5 terms. Or
when you think about, when we use the binomial expansion,
and you might want to watch some videos on that.
There you also have, if you're taking it to the nth power,
you end up with n plus 1 terms.
So in this example, we have a binomial-- a binomial just
means a two-term polynomial.
And we're taking it to the twentieth power.
If we were taking it to third power, we'd have four terms.
If we were taking it to the fourth power, we'd have 5
terms. So if we're taking it to the twentieth power, we're
going to have 21 terms. And that's choice B.
Next problem.
All right, they want us to keep doing these and we'll do
it because they want us to.
All right, what are the first four terms of the expansion of
1 plus 2x to the sixth power?
And here we'll just use the binomial expansion because I
think that's what they want us to do.
And that's probably the fastest way to do it.
So let's just think of the coefficients first. So this is
going to be-- well let's just do it.
This is going to be equal to 6 choose 0.
We're just going to do the first four terms. Times 1 to
the 6 times 2x to the 0-- so I don't have to write that--
plus 6 choose 1.
And actually, we'll probably just have to figure out the
third term because that's the only time where
they start to change.
6 choose 1 times 1 to the fifth power times 2x to the
first power, times 2x.
Plus 6 choose 2 times-- I mean these 1s, it doesn't matter
what power it is-- 1 to the fourth times 2x squared plus--
let's see, they want the first so let's do the last one-- 6
choose 3 times 1 to the third-- you have to decrement
the 1 every time-- times 2x to the third.
All right, and if we look at the choices,
just to save time.
If we look at a choices, OK, we agree that the first term
is going to be 1.
So that, we can definitely say is going to be 1 because all
the choices are 1.
The second term is definitely going to be 12x.
And then we start getting some disagreement
on the third term.
So let's see what the third term.
6 choose to that's equal to 6 back toward ill over to affect
or you over fix minus is too fact or you scroll down a
little bit so that is equal to 6 factorial, which is 6 times
5 times 4 times 3 times 2 times 1.
But the 1 doesn't change anything.
Divided by 2 factorial, which is 2 times 1, or just 2.
Divided by 6 minus 2 factorial, that's 4 factorial.
Royal So times 4 times 3 times 2 times 1.
That cancels out with that.
The 2 and the 6 cancel out.
You get 3.
So you end up with 3 times 5.
So the 6 choose 2 becomes 15, 3 times 5, that's 15.
Times 1 to the fourth, which is 1, times 2x squared.
So times 4x squared.
And so that's 15 times 4, that's 60x squared.
So the third term is going to be 60x squared.
And we're done.
Choice D is the only one that has 60x squared
as the third term.
So we're done.
We didn't even have to go to the fourth term.
So, by carefully looking at the choices we actually only
have to evaluate one of the terms.
Next problem.
Let's see.
All right.
Problem 69.
They say-- let me copy and paste it-- what is the sum of
the infinite geometric series, 1/2 plus 1/4 plus 1/8 all the
way up there.
I'll be frank.
I do now remember the formula.
But a lot of times in life you might forget the formula.
And there's a trick is to re-proving
the formula for yourself.
Just in case you forget it when you're 32
years old like me.
So let's say that I'm trying to find the infinite sum of a
geometric series.
So let's say that the sum-- and I'm going to re-prove it
right here in the midst of this test. And I've done this
in midst of tests and it's saved me.
So it's nice to know the proof.
And it actually impresses people, if you know people who
are impressed by proofs.
So let's say the sum is equal to a to the 0-- well actually
let's say-- well, see this is interesting.
Because they're starting at 1/2.
Actually let's just do the concrete numbers.
Because we want to figure out this exact sum.
So let's say that the sum is equal to 1/2 to the 1 1/2
squared plus 1/2 to the third-- it's the whole 1/2 to
the third-- plus-- and you just keep going.
Fair enough.
Now let's take 1/2 times the sum.
So 1/2 times this.
What is this equal to?
Well if I took 1/2 times this whole thing, now 1/2, I would
have to distribute it over each of these terms. So 1/2
times this term, times this first term, 1/2 times 1/2 to
the first, well that's now going to be 1/2 squared.
1/2 squared.
All I did is I'm distributing this 1/2 times each of the
terms. 1/2 times 1/2 to the first. That's 1/2 squared.
Now 1/2 times 1/2 squared, well that's going to be 1/2 to
the third power.
And it just keeps going.
It's an infinite series.
So let me ask you something.
If I subtracted this from that, what do I get?
Let's see what I get.
So I get S minus 1/2 S-- I'm subtracting that from that--
is equal to this minus this.
Well if I'm subtracting the green stuff from the yellow
stuff, this is going to cancel out with that, that's going to
cancel out with that.
And all I'm going to be left with is this
first thing right here.
That's the trick.
And you could prove it generally for any base.
And I thought it was interesting because a lot of
times they start with the 0th power.
And this problem was interesting because they start
with 1/2 to the first power.
So anyway let's figure it out.
S minus 1/2 S, well that's just 1/2 S.
1/2 S is equal to 1/2.
And then you have S is-- divided both sides by 1/2 or
multiply both sides by 2 and you get-- S is equal to 1.
Now, a lot of you may know there is a formula.
Let me tell you what the formula says.
The formula that you may or may not have memorized, is if
you start at n is equal to 0 and if you go to infinity of a
to the n-- in this case a is 1/2-- this is going to be
equal to 1 over 1 minus a.
So your natural reaction would have been, oh OK, well that
sum that we saw, that's equal to the sum of n is equal to 0
to infinity of 1/2 to the n, which would be equal to 1 over
1 minus 1/2.
What's that?
1 minus 1, that's 1 over 1/2, which would be equal to 2.
And you'd be like, my God, what did I do wrong there?
And I'll show you what you did wrong.
This sum that you took the sum of-- remember n is starting at
0-- so this sum that sums up to 2 is 1/2 to 0 is 1 plus 1/2
to the 1, 1/2, plus 1/2 squared, 1/4, plus 1/8, so on
and so forth.
That is equal to 2.
This problem-- and that's why it was a little tricky-- this
problem, they didn't want this sum, they wanted this sum.
1/2 plus 1/4 plus 1/8 plus, so forth and so on.
So they wanted everything but this 1.
So if you subtracted this 1 from both sides then you would
say, oh this sum must be equal to 1, which is the answer we
got that way.
So that's why sometimes, memorizing formulas, this
formula's a nice formula to memorize, it's very simple.
But it sometimes becomes really confusing when you're
like, oh boy, but this is when n starts 0 but what happens
when n starts at 1.
And to some degree I really like this problem because it's
testing you to see if you really understand what the
formula's all about.
Anyway, I'll see you in the next video.