How do you find $S_n$ for the geometric series $a_1=4$ , r=-3, n=5?

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Answer 1 · 2016-10-18T17:16:33

$S_5=244$

Here in given geometric series $a_1=4$ and $r=-3$

Now if $a_1=a$ , $S-n$ the sum of geometric series is

$S_n=a+ar+ar^2+.................+ar^(n-1)$ and therefore

$rS_n=ar+ar^2+.................+ar^(n-1)+ar^n$

and subtracting first from second $S_n(r-1)=a(r^n-1)$

and $S_n=(a(r^n-1))/(r-1)$

Hence in the given geometric series

$S_5=(4((-3)^5-1))/(-3-1)$

= $(4(-243-1))/(-4)$

= $244$

How do you find S_nS_n for the geometric series a_1=4a_1=4, r=-3, n=5?