Worked example: finite geometric series (sigma notation)
About this video
The video applies the formula S_n = a(1 - r^n)/(1 - r) to the series written in sigma notation as the sum from k = 0 to 99 of 2 x 3^k. By expanding the first few terms it identifies a = 2, r = 3 and n = 100 (not 99, since k starts at 0), giving 3^100 - 1.
You'll be able to
- identify the first term, common ratio and number of terms from sigma notation.
- count terms correctly when the index starts at 0.
- apply the finite geometric series formula and simplify the result.
Key words
- Finite geometric series
- The sum of a fixed number of terms of a geometric sequence.
- S_n
- The sum of the first n terms of a series.
- Common ratio
- The number each term is multiplied by to get the next term.
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.
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