How do you find the sum of the geometric series $Sigma 2(-3)^(n-1)$ n=1 to 6?

Question

2069 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-02-06T12:00:20

$= - 364$

The series is: $S = Sigma_(n = 1)^(6) 2(-3)^(n-1) = 2 Sigma_(n = 1)^(6) (-3)^(n-1)$

$= 2( (-3)^0 + (-3)^1 + ....)$

Which can be written as:

$S = 2( 1 + r + .... + r^5)$ , with $r = -3$

$implies r S = 2( r + r^2 + .... + r^6)$

$implies rS - S = (r-1) S = 2( r^6 -1)$

$implies S = 2( r^6 -1)/(r-1)$

$= 2( (-3)^6 -1)/((-3)-1) = - 364$

How do you find the sum of the geometric series Sigma 2(-3)^(n-1)Sigma 2(-3)^(n-1) n=1 to 6?