How do you find sum of geometric series 2+6+18+...+1458?

1 Answer
Mar 15, 2016

2186

Explanation:

Given that first term a_1=2

Common ratio q=a_2/a_1=6/2=3

General expression for nth term is

a_n=a_1*q^(n-1)

To ascertain what is number of the last term, insert in the above equation

1458=2*3^(n-1)
=> 3^(n-1) = 1458/2=729
Writing 729 as power of 3 we obtain

3^(n-1) = 3^6, comparing the exponents
n-1=6
or n=7
Now sum of n terms is given by the expression

S_n = (a_1(1-q^n))/(1-q)

S_7 = (2 (1-3^7)) / (1-3)
= (2 (1-2187))/-2 = 2186