How do you find the sum of the finite geometric sequence of $Sigma 3(3/2)^n$ from n=0 to 20?

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Answer 1 · 2018-03-28T00:41:18

$sum_(n=0)^(20)3(3/2)^n~~29921.31057$

We are asked to evaluate

$sum_(n=0)^(20)3(3/2)^n=3+3(3/2)+3(3/2)^2+3(3/2)^3+cdots+3(3/2)^20$

This is a geometric series of the form

$sum_(n=0)^kaR^n$

with $a=3$ and $R=3/2$

There is a formula to calculate the $k$ th partial sum of a geometric series. It is given by:

$S_k=a((1-R^(k+1))/(1-R))$

Therefore:

$S_20=a((1-R^21)/(1-R))=3((1-(3/2)^21)/(1-3/2))$

$=3((1-(3/2)^21)/(-1/2))=-6(1-(3/2)^21)=6((3/2)^21-1)$

Obviously this is going to be a pretty big number. If you punch it into a calculator, you'll find it's

$~~29921.31057$

How do you find the sum of the finite geometric sequence of Sigma 3(3/2)^nSigma 3(3/2)^n from n=0 to 20?