How do you find the sum of the geometric series $Sigma 64(3/4)^(n-1)$ n=1 to 8?

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Answer 1 · 2018-06-27T14:41:12

$S_8=(4^8-3^8)/4^7$

The sum of the first $n$ terms of a geometric series with general term $b_m$ and common ratio $q$ is

$S_n = b_1 * (q^n-1)/(q-1)$

For $q<1$ , this is usually written as

$S_n = b_1 (1-q^n)/(1-q)$

$b_n = 64(3/4)^(n-1)=> {(b_1 = 64),(q=b_(n+1)/b_n = 3/4) :}$

$S_8 = 64*(1-(3/4)^8)/(1-3/4)=(4^8-3^8)/4^7$

How do you find the sum of the geometric series Sigma 64(3/4)^(n-1)Sigma 64(3/4)^(n-1) n=1 to 8?