Question

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

Answer 1

The sum of geometric sequence:
S n = a 1 * ( q^n - 1 ) / ( q - 1 )
where:  a 1 = 1,
q = a 2 : a 1 = 4 : 1 = 4
S 8 = 1 * ( 4^(8) - 1 ) / ( 4 - 1 ) = 
= ( 65,536 - 1 ) / ( 4 - 1 ) = 65,535 / 3 = 21,845

Answer 2

Answer:

The the sum of the geometric sequence 1, 4, 16, … up to 8 terms is 21845.

Step-by-step explanation:

Given the geometric series 1, 4, 16, …

we have to find the sum of the GP if there are 8 terms.

Series:  1, 4, 16, …

First term, a=1

Number of terms, n=8

Common ratio, $r=\frac{4}{1}=\frac{16}{4}=4$

The formula to find the sum of above G.P is

$S_n=\frac{a(r^n-1)}{r-1}$

     $=\frac{1(4^8-1)}{4-1}=\frac{65536-1}{3}=\frac{65535}{3}=21845$

Hence, the the sum of the geometric sequence 1, 4, 16, … up to 8 terms is 21845.