How do you find $S_{n}$ $S_n$ for the geometric series $a_{1} = 162$ $a_1=162$ , r=1/3, n=6?

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Answer 1 · 2016-12-01T07:11:25

The answer is $= \frac{2186}{9}$ $=2186/9$

Let $S_{n}$ $S_n$ be the sum of a geometric series

$S_n=a_1+a_1r+ ........ a_1r^n$

Then, $rS_n=a_1r+..................+a_1r^(n+1)$

Therefore,

$S_n(1-r)=a_1(1-r^(n+1))$

$S_n=(a_1(1-r^(n+1)))/(1-r)$

$a_1=162$

$r=1/3$

$n=6$

$S_6=162(1-(1/3)^7)/(1-1/3)$

$=162*3/2(1-(1/3)^7)$

$=243-243/3^7$

$=243-1/9$

$=2186/9$

How do you find S_nSnS_n for the geometric series a_1=162a1=162a_1=162, r=1/3, n=6?