What I hope to do in this video is derive
a formula for the sum of a finite geometric series.
And right here is a finite geometric series.
I start with some first term a.
Notice, I wrote it as a times r to the 0.
But r to the 0-- for any r.
We're assuming r is non-zero.
This is going to be 1 right over here.
So our first term is going to be a.
And then each successive term is a previous term
times our common ratio, times r.
So a times r is a times r to the first.
That times r is ar squared.
On and on, all the way until we go a times r to the n.
This is a finite series right over here.
So let's just define this series.
I'll use the letter S sub n as being
equal to all of this business right over here.
Now, let me multiply S sub n times our common ratio.
So let me write this right over here.
So let's say we have r-- I'm going
to do the same color as our previous r-- r times S sub n.
And what is that going to be equal to?
Well, we're just going to multiply each term here by r.
And we could even multiply this thing by r if we want as well.
This would now be the sum from k equals 0 to n of a times r.
But now, instead of r to the k, we're
now multiplying each of these terms times r again.
So it's going to be the k plus 1 power.
Or-- and it's a little easier to conceptualize right over here--
let's just multiply each of these terms times r.
So a times r to the 0 times r is going
to be ar to the first power.
So that's going to be a times r to the first power.
Let me write that down. a times r to the first power.
Or just a times r because, obviously, r to the first
is just r.
And then, plus-- let's multiply this term times r.
So plus a times r squared.
And we'll just keep on going.
We're going to get all the way to a times r to the n plus 1.
So let me write it like this.
So we're going to go to plus-- and typically, you'd
only write three dots.
I went overboard with my dots.
So let me do it-- dot, dot, dot.
So plus a times r to the n plus a times r to the n plus 1.
So notice, I just took each of these terms
and multiplied them by r.
You see that right over here.
Multiply by r, get that term.
And then, I don't show the other ones multiplied by r,
but all the way to multiply this one by r.
And you get this term right over here.
Now, what I want to do is something really neat.
And this is one of my favorite derivations
in all of mathematics because you're
able to do something pretty neat at this stage.
What happens if I were to subtract r times S sub n
from S sub n?
So let's think about that.
Well, over here, we would just get S sub n minus r times S sub
n.
So notice, all I'm doing is I'm starting with this
and I'm subtracting the same thing
from both sides of the equation.
On the left-hand side, I'm subtracting r times S sub n.
And on the right-hand side, from this
I'm going to subtract all of this.
So what do we get on the right-hand side?
So this is going to be equal to--
and I'm going to ignore this for now
because what we really care about is this business.
So a times r.
What I'm going to do is I'm just going to cancel out
all of the terms that have the same degree.
So for example, a times r to the 0.
There's no other r to the 0 term here.
So we're just going to have that a times r to the 0.
Or I could just write that as a, obviously. r to the 0
is going to be 1.
So I'll just write that as a.
That's that right over there.
And then, we have a times r to the first power
minus r to the first power.
Well, that's just going to cancel out.
These are going to cancel out.
They're going to keep canceling out.
They're going to keep canceling out all the way
until all we have left over is this.
And there's nothing to subtract this from.
We have no a times r to the n plus 1 power up here.
So we're just going to subtract it.
So minus a times r to the n plus 1 power.
And now we can just solve for our sum.
Remember, that's the whole point of this video
to begin with-- to try to derive a formula for this sum.
So let's factor out S sub n here.
So this is S sub n times 1 minus r
is equal to this right over here.
It's equal to the first term minus the first term times r
to the n plus 1.
So one more power times the number
of terms in our original sum.
So let's do that-- times r to the n plus 1 power.
And so to solve for S sub n, we just
have to divide both sides by 1 minus r.
And so we are left with-- if we assume r is not equal to 1
in this case-- we are left with S sub n is equal to a minus
a times r-- I'm just rewriting what I have up here-- r
to the n plus 1 over 1 minus r.
And you say, hey, Sal.
Well, why is this at all interesting?
Why is this useful?
Well, let's actually apply this formula
that we just finished deriving.
Let's say that we have a geometric series.
So let's say our first term is-- oh, I don't know.
Let's say our first term is 3.
I'll do the same colors as I was doing before.
It is 3 times 1/2 to the k-th power.
And we're going to start at k equals 0 all the way to-- I
don't know-- k equals 100.
And just to visualize what this looks like,
the first term is just going to be 3 times
one half to the 0 power.
It's just 3.
So our first term is 3, as was intended.
So our first term is 3.
Plus 3 times 1/2 to the first power now.
This is 3 times 1/2 to the 0-th power.
Now, 3 times 1/2 to the first power.
3 times 1/2.
I could write a first power there if I'd like.
And then, plus 3 times 1/2 half squared.
And we'll just keep on going all the way-- I'll do it
right over here-- plus, we're going
to keep on going all the way until we get to 3 times 1/2
to the 100th power.
Now, you could imagine you could probably
calculate this with a calculator,
but it would get very, very hard.
And it would be very, very time-consuming
to add up these 101 terms right over here.
Instead, we can apply this formula.
So this sum right over here is going
to be equal to-- if we were to add it up--
it's going to be equal to our first term, which is 3.
It's going to be equal to 3 minus 3 times 1/2-- 1/2
is our common ratio here-- times 1/2 to the n plus 1.
Well, n is 100.
So it's to the 101 power.
All of that over 1 minus 1/2.
And now this is much easier.
It would be hard to do it in your head,
but this is much easier to type into a calculator
and actually get a reasonable answer.
So let's actually do that.
It'll feel satisfying.
So let's clear this.
So it's going to be 3 minus 3 times-- I could write 1/2
as 0.5-- 0.5 to the 101 power.
Did I get that [INAUDIBLE] right?
3 minus 3 times 0.5 to the 101 power divided by-- well,
1 minus 1/2 is just 1/2-- so it's just
going to be divided by 1/2 or divided by 0.5.
And we get 6.
We get 6.
And it's not going to be exactly 6.
And we're going to see that, when
we take the infinite sum, that's going to be exactly 6.
But it's going to be pretty darn close.
We're actually, we're hitting up against the precision
of the calculator.
So based on the precision of the calculator,
we get, it's going to be pretty close to 6.