What is the sum of the geometric sequence 1, 4, 16, … if there are 8 terms?

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Answer 1 · 2016-01-25T03:45:36

$a_2/a_1=4/1=4$

$a_3/a_2=16/4=4$

$implies$ common ratio $=r=4$ and $a_1=1$

Sum of geometric series is given by

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

Where $n$ is number of terms, $a_1$ is the first term, $r$ is the common ratio.

Here $a_1=1$ , $n=8$ and $r=4$

$implies Sum=(1(1-4^8))/(1-4)=(1-65536)/-3=(-65535)/-3=21845$

Hence, the sum is $21845$