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

Question

20684 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-03-15T08:02:03

$2186$

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

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

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