WEBVTT

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We're on problem 67.

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They ask us how many terms of
the binomial expansion of x

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squared plus 2y to
the third to the

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twentieth power contain?

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And so right when you see, oh
my god, to the twentieth

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power, that will take me forever
to do, and then I'll

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have to count the terms. But
remember, they're just asking

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you how many terms.
They're not asking

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you to find the expansion.

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So you just have to say, oh boy,
let me just think about

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this a little bit.

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And think of it this way: x plus
y to the first power has

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how many terms?

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It has-- well it's
just x plus y.

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And so it has 2 terms.
x plus y, squared.

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That is equal to x squared
plus 2xy plus y squared.

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So that has 3 terms. If I were
to do x plus e The third

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power, I could do a binomial
expansion, I could do it with

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Pascal's Triangle.

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Let me do it with Pascal's
Triangle.

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I have the 1 and the 1.

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That's to the first power.

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The second power, those
are the coefficients.

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The third power, the
coefficients become 1, 3--

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because 1 plus 2 is 3-- 3, 1.

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And actually I don't even
have to expand it out.

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Ill do it just for a
little practice.

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It x to the third plus 3x
squared y plus 3xy squared

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plus y to the third.

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I just used these.

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But when you take any binomial
and you take it to the third

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power, you end up with 4 terms.
And depending on how

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you think of it, if you think of
it with Pascal's Triangle,

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every time we go down we're
just going to be adding

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another term.

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When we take the fourth power
it's going to have 5 terms. Or

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when you think about, when we
use the binomial expansion,

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and you might want to watch
some videos on that.

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There you also have, if you're
taking it to the nth power,

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you end up with n
plus 1 terms.

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So in this example, we have a
binomial-- a binomial just

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means a two-term polynomial.

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And we're taking it to
the twentieth power.

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If we were taking it to third
power, we'd have four terms.

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If we were taking it to the
fourth power, we'd have 5

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terms. So if we're taking it to
the twentieth power, we're

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going to have 21 terms.
And that's choice B.

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Next problem.

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All right, they want us to keep
doing these and we'll do

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it because they want us to.

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All right, what are the first
four terms of the expansion of

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1 plus 2x to the sixth power?

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And here we'll just use the
binomial expansion because I

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think that's what they
want us to do.

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And that's probably the
fastest way to do it.

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So let's just think of the
coefficients first. So this is

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going to be-- well
let's just do it.

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This is going to be equal
to 6 choose 0.

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We're just going to do the first
four terms. Times 1 to

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the 6 times 2x to the 0-- so I
don't have to write that--

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plus 6 choose 1.

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And actually, we'll probably
just have to figure out the

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third term because that's
the only time where

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they start to change.

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6 choose 1 times 1 to the fifth
power times 2x to the

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first power, times 2x.

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Plus 6 choose 2 times-- I mean
these 1s, it doesn't matter

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what power it is-- 1 to the
fourth times 2x squared plus--

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let's see, they want the first
so let's do the last one-- 6

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choose 3 times 1 to the third--
you have to decrement

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the 1 every time-- times
2x to the third.

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All right, and if we look
at the choices,

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just to save time.

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If we look at a choices, OK, we
agree that the first term

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is going to be 1.

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So that, we can definitely say
is going to be 1 because all

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the choices are 1.

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The second term is definitely
going to be 12x.

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And then we start getting
some disagreement

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on the third term.

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So let's see what
the third term.

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6 choose to that's equal to 6
back toward ill over to affect

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or you over fix minus is too
fact or you scroll down a

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little bit so that is equal to
6 factorial, which is 6 times

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5 times 4 times 3
times 2 times 1.

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But the 1 doesn't
change anything.

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Divided by 2 factorial, which
is 2 times 1, or just 2.

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Divided by 6 minus 2 factorial,
that's 4 factorial.

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Royal So times 4 times
3 times 2 times 1.

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That cancels out with that.

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The 2 and the 6 cancel out.

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You get 3.

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So you end up with 3 times 5.

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So the 6 choose 2 becomes 15,
3 times 5, that's 15.

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Times 1 to the fourth, which
is 1, times 2x squared.

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So times 4x squared.

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And so that's 15 times 4,
that's 60x squared.

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So the third term is going
to be 60x squared.

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And we're done.

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Choice D is the only one
that has 60x squared

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as the third term.

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So we're done.

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We didn't even have to go
to the fourth term.

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So, by carefully looking at the
choices we actually only

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have to evaluate one
of the terms.

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Next problem.

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Let's see.

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All right.

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Problem 69.

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They say-- let me copy and paste
it-- what is the sum of

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the infinite geometric series,
1/2 plus 1/4 plus 1/8 all the

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way up there.

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I'll be frank.

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I do now remember the formula.

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But a lot of times in life you
might forget the formula.

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And there's a trick
is to re-proving

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the formula for yourself.

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Just in case you forget
it when you're 32

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years old like me.

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So let's say that I'm trying to
find the infinite sum of a

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geometric series.

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So let's say that the sum-- and
I'm going to re-prove it

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right here in the midst of this
test. And I've done this

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in midst of tests and
it's saved me.

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So it's nice to know
the proof.

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And it actually impresses
people, if you know people who

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are impressed by proofs.

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So let's say the sum is equal to
a to the 0-- well actually

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let's say-- well, see
this is interesting.

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Because they're starting
at 1/2.

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Actually let's just do
the concrete numbers.

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Because we want to figure
out this exact sum.

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So let's say that the sum is
equal to 1/2 to the 1 1/2

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squared plus 1/2 to the third--
it's the whole 1/2 to

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the third-- plus-- and
you just keep going.

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Fair enough.

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Now let's take 1/2
times the sum.

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So 1/2 times this.

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What is this equal to?

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Well if I took 1/2 times this
whole thing, now 1/2, I would

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have to distribute it over each
of these terms. So 1/2

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times this term, times this
first term, 1/2 times 1/2 to

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the first, well that's now
going to be 1/2 squared.

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1/2 squared.

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All I did is I'm distributing
this 1/2 times each of the

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terms. 1/2 times 1/2 to the
first. That's 1/2 squared.

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Now 1/2 times 1/2 squared, well
that's going to be 1/2 to

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the third power.

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And it just keeps going.

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It's an infinite series.

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So let me ask you something.

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If I subtracted this from
that, what do I get?

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Let's see what I get.

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So I get S minus 1/2 S-- I'm
subtracting that from that--

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is equal to this minus this.

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Well if I'm subtracting the
green stuff from the yellow

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stuff, this is going to cancel
out with that, that's going to

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cancel out with that.

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And all I'm going to be
left with is this

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first thing right here.

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That's the trick.

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And you could prove it generally
for any base.

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And I thought it was interesting
because a lot of

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times they start with
the 0th power.

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And this problem was interesting
because they start

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with 1/2 to the first power.

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So anyway let's figure it out.

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S minus 1/2 S, well
that's just 1/2 S.

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1/2 S is equal to 1/2.

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And then you have S is-- divided
both sides by 1/2 or

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multiply both sides by 2 and
you get-- S is equal to 1.

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Now, a lot of you may know
there is a formula.

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Let me tell you what
the formula says.

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The formula that you may or may
not have memorized, is if

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you start at n is equal to 0 and
if you go to infinity of a

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to the n-- in this case a is
1/2-- this is going to be

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equal to 1 over 1 minus a.

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So your natural reaction would
have been, oh OK, well that

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sum that we saw, that's equal to
the sum of n is equal to 0

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to infinity of 1/2 to the n,
which would be equal to 1 over

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1 minus 1/2.

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What's that?

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1 minus 1, that's 1 over 1/2,
which would be equal to 2.

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And you'd be like, my God, what
did I do wrong there?

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And I'll show you what
you did wrong.

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This sum that you took the sum
of-- remember n is starting at

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0-- so this sum that sums up to
2 is 1/2 to 0 is 1 plus 1/2

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to the 1, 1/2, plus 1/2 squared,
1/4, plus 1/8, so on

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and so forth.

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That is equal to 2.

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This problem-- and that's why it
was a little tricky-- this

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problem, they didn't want this
sum, they wanted this sum.

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1/2 plus 1/4 plus 1/8 plus,
so forth and so on.

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So they wanted everything
but this 1.

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So if you subtracted this 1 from
both sides then you would

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say, oh this sum must be equal
to 1, which is the answer we

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got that way.

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So that's why sometimes,
memorizing formulas, this

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formula's a nice formula to
memorize, it's very simple.

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But it sometimes becomes really
confusing when you're

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like, oh boy, but this is when
n starts 0 but what happens

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when n starts at 1.

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And to some degree I really like
this problem because it's

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testing you to see if you really
understand what the

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formula's all about.

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Anyway, I'll see you
in the next video.

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