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## Séries géométriques finies

Current time:0:00Total duration:9:38

# DEPRECATED Finite geometric series

## Video transcript

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. 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.