What is the sum of the geometric sequence 4, 16, 64, … if there are 8 terms?

Solution:

Sum of the geometric sequence is :

S = a + ar + ar¹ + ar² +....+ ar^{n - 1}

First term of the series a is 4.

Common ratio  is r.

To find r,

r = 16/4

r = 4

Since r > 1, sum of geometric sequence can be found by using the relation,

S_n = a(rⁿ - 1)/(r - 1), r ≠ 1

Given, n = 8

S₈ = 4(4⁸ - 1)/(4 - 1)

= 4(65536 - 1)/3

= 4(65535)/3

= 87380

Therefore , the sum of the geometric sequence is 87380.

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What is the sum of the geometric sequence 4, 16, 64, … if there are 8 terms?

Summary:

The sum of the geometric sequence 4, 16, 64 ... if there are 8 terms is 87380.