How do you find the sum of the geometric series $Sigma 1/4*2^(n-1)$ n=1 to 5?

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Answer 1 · 2016-12-16T06:14:02

Sum of first five terms $n=1$ to $n=5$ of given series is $7 3/4$

In the geometric series $Sigma 1/4*2^(n-1)$

first term $a_1$ is $a_1=1/4xx2^(1-1)=1/4xx1=1/4$

Subsequent terms are $a_2=1/4xx2^(2-1)=1/4xx2=1/2$

and $a_3=1/4xx2^(3-1)=1/4xx4=1$

Observe that the power of $2$ increases by one from a term to the next term,

Hence common ration $r=2$

As sum of a geometric series $S_n$ , with first term as $a_1$ and common ratio as $r$ is

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

Sum of first five terms $n=1$ to $n=5$ of given series is

$S_5=(1/4(2^5-1))/(2-1)=1/4xx(32-1)/1=1/4xx31=31/4=7 3/4$

How do you find the sum of the geometric series Sigma 1/4*2^(n-1)Sigma 1/4*2^(n-1) n=1 to 5?