How do you find the sum of the infinite geometric series given $Sigma 40(3/5)^(n-1)$ for n=1 to $oo$ ?

Question

$sum_(n=1)^oo 40(3/5)^(n-1)$

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Answer 1 · 2018-04-15T02:16:38

$100$

The sum of an infinite geometric series in the below form is given by

$sum_(n=1)^ooa(r)^(n-1)=a/(1-r),$ so long as $-1<r<1.$

Here, for $sum_(n=1)^oo40(3/5)^(n-1), |r|=3/5<1$ , so it will have a finite sum. Also $a=40,$ so

$sum_(n=1)^oo40(3/5)^(n-1)=40/(1-3/5)=40/(2/5)=40*5/2=100$