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## Geometric series (with summation notation)

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# Worked example: finite geometric series (sigma notation)

CCSS Math: HSA.SSE.B.4

## Video transcript

- [Voiceover] Let's do some examples where we're finding sums
of finite geometric series. Now let's just remind
ourselves in a previous video we derived the formula where
the sum of the first n terms is equal to our first term times one minus our common ratio to the nth power all over
one minus our common ratio. So let's apply that to this
finite geometric series right over here. So what is our first term
and what is our common ratio? And what is our n? Well, some of you might
just be able to pick it out by inspecting this here, but
for the sake of this example, let's expand this out a little bit. This is going to be equal to
two times three to the zero, which is just two, plus two times three to the first power, plus two times three to the second power, I can write first power there, plus two times three to the third power, and we're gonna go all
the way to two times three to the 99th power. So what is our first term? What is our a? Well, a is going to be two. And we see that in all
of these terms here. So a is going to be two. What is r? Well, each successive term,
as k increases by one, we're multiplying by three again. So, three is our common ratio. So that right over there, that is r. Let me make sure that we, that is a. And now what is n going to be? Well, you might be tempted to say, well, we're going up to k
equals 99, maybe n is 99, but we have to realize that
we're starting at k equals zero. So there is actually 100 terms here. Notice, when k equals zero,
that's our first term, when k equals one, that's our second term, when k equals two, that's our third term, when k equals three,
that's our fourth term, when k equals 99, this is
our 100th term, 100th term. So what we really want
to find is S sub 100. So let's write that down, S sub 100, for this geometric series
is going to be equal to two times one minus
three to the 100th power, to the 100th power, all of that, all of that over, all of
that over one minus three. And we could simplify
this, I mean at this point it is arithmetic that
you'd be dealing with, but down here you would
have a negative two, and so you'd have two
divided by negative two so that is just a negative. And so negative of one
minus three to the 100th, that's the same thing, this is
equal to three to the 100th, three to the 100th power minus one. And we're done.