Algebra II: Binomial Expansions, Geometric Series Sum

67-69, binomial expansions and the sum of a geometric series

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.