infinite geometric series

[hgaon001]

It says find the infinite geometric series from n=2 to intinity of 34[(-1/4)^n]

im not sure how to start i know i'll use a/1-r but since it starts at n=2 im not sure how 2 approach the problem

 

[BigGlenntheHeavy]

Convergence of a geometric series:

$\displaystyle \sum_{n=0}^{\infty} ar^{n} \ = \ \frac{a}{1-r}, \ 0<|r|<1.$

$\displaystyle Hence, \ \sum_{n=2}^{\infty} 34(\frac{-1}{4})^{n} \ = \ 34[\frac{1}{16}-\frac{1}{64}+\frac{1}{256}-\frac{1}{1024}+,,,,]$

$\displaystyle Now \ a \ =\frac{1}{16} \ and \ r \ = \ \frac{-1}{4}, \ ergo \ Sum \ = \ 34\bigg[\frac{\frac{1}{16}}{1-(\frac{-1}{4})}\bigg]$

$\displaystyle = \ 34\bigg[(\frac{1}{16})(\frac{4}{5})\bigg] \ = \ \frac{17}{10}.$

Therefore, the series converges to 17/10.