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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?
• 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.
• Does the k of the sigma notation equal the n of the equation?
• 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!
• 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.
• 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.
• At (this is a suggestion), why couldn't you have solved 3^100-1?
• 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.
• Why does 2*3 to the 0th power equal to 2? Shouldn't it be zero?
• I'm confused about what happens to the negative at
• 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
• 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?
• Yes you can, that's called 'reindexing'. Neither is preferred, there may be contexts where you want one form or the other.
• What about infinite geometric sequences? What is the formula for the sum?
• If the sequence is S=a+ar+ar²+ar³+...
Multiply by the common ratio:
rS=ar+ar²+ar³+ar⁴+...

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)
• I'm confused does the negative from the denominator cancel out as well?