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### Course: Calculus, all content (2017 edition) > Unit 7

Lesson 4: Finite geometric series# Worked example: finite geometric series (sigma notation)

Sal evaluates the geometric series Σ2(3ᵏ) for k=0 to 99 using the finite geometric series formula a(1-rⁿ)/(1-r).

## Want to join the conversation?

- My professor handed me a sheet listing the different formulas for geometric series and it shows up as Sn = a*(r^n -1) / r-1

Will this give a different answer than your formula of Sn = a*(1-r^n) / 1-r?(10 votes)- The two formulas are equivalent. If you multiply Sn = a*(r^n-1) / r-1 by -1 / -1, it will not change the value because -1 / -1 = 1.

a*(r^n-1) * -1 = -a*(r^n-1) = a*(-r^n+1) = a*(1-r^n).

r-1 * -1 = -r+1 = 1-r.

Thus the formula becomes Sn = a*(1-r^n) / 1-r, which means the two formulas are equivalent.(23 votes)

- Does the k of the sigma notation equal the n of the equation?(8 votes)
- Good question!

No, it doesn't, and it's important to understand the difference.

We're talking about 2 things - the "S sub n" equation and the sigma notation of the series.

The "n" in the"S sub n" equation only represents how many terms there are all together in the series.

So, "S sub 100" means the sum of the first 100 terms in the series.

The k of the sigma notation tells us what needs to be substituted into the expression in the sigma notation in order to get the full series of terms.

So, if k goes from 0 to 99, there are 100 terms, so 100 would be used as "n" in the "S sub n" equation.

If k goes from 3 to 24, there are 22 terms, so 22 would be used as "n" in the "S sub n" equation.

(If desired, the individual terms of the series could be found by substituting each of the "k" values into the sigma notation expression.)

Hope this helps -- even if only in a small way!(24 votes)

- There is no mention of the formula for geometric series in the previous videos. It is just introduced here as though it should be something already learned. It is only explained in the "Finite geometric series formula justification" which is the last video in this section.(14 votes)
- I agree. No there is no mention of it.It is like you say,it is only explained later on the last section of the tutorial. It should be corrected as it is confusing and not typical of the lessons where every step follows the previous one.(12 votes)

- At2:44(this is a suggestion), why couldn't you have solved 3^100-1?(5 votes)
- We could but it's difficult to do without a calculator, because 3¹⁰⁰ (that is 3 times itself 100 times) is a very huge number. My calculator shows the result a number that is around 40 digits. Even with calculator, most regular calculators wouldn't be able to display all of its digits.(7 votes)

- Why does 2*3 to the 0th power equal to 2? Shouldn't it be zero?(3 votes)
- To get 0, you need to be multiplying by 0. You aren't.

3^0 = 1, not 0.

then multiply 2*1 and you get 2.

See this video on exponent of 0: https://www.khanacademy.org/math/cc-sixth-grade-math/x0267d782:cc-6th-exponents-and-order-of-operations/x0267d782:powers-of-whole-numbers/v/the-zeroth-power(3 votes)

- I'm confused about what happens to the negative at2:37(3 votes)
- you have (2*(1-3^100))/(-2).

then the 2s cancel out: (1-3^100)/(-1).

negative from the -1 goes to the top: -(1-3^100)/1.

1 in denominator goes away: -(1-3^100).

distribute negative: -1+3^100

rewrite order: 3^100-1(2 votes)

- Could we write
`Σ2(3^(k-1)) for k=1 to 100`

instead of`Σ2(3^k) for k=0 to 99`

? What's the preferred way of writing this?(2 votes)- Yes you can, that's called 'reindexing'. Neither is preferred, there may be contexts where you want one form or the other.(3 votes)

- What about infinite geometric sequences? What is the formula for the sum?(2 votes)
- If the sequence is S=a+ar+ar²+ar³+...

Multiply by the common ratio:

rS=ar+ar²+ar³+ar⁴+...

Add a:

rS+a=a+ar+ar²+ar³+...

Now the right-hand side is the original S:

rS+a=S

Solve for S:

a=S-rS=S(1-r)

S=a/(1-r)(3 votes)

- I'm confused does the negative from the denominator cancel out as well?(2 votes)
- No. Sal divides 2/(-2) = -1. He puts a "-" in front of the parentheses as the -1.

Hope this helps.(2 votes)

- So n in this context is the number of whole numbers the finite geometric series has ?(1 vote)
- Yes, I think so. In the case in the video, n = 100(3 votes)

## 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.