Question

how is (1+(2^n))/(3^n) rewritten into the standard geometric series notation?

ar^(n-1)

Answer 1

Hi pgarcia164, I was flicking through old questions and saw this unanswered.

You need to put your summand in the following form:\[\frac{1+2^n}{3^n}=\frac{1}{3^n}+\frac{2^n}{3^n}=\frac{1}{3^n}+\left( \frac{2}{3} \right)^n\]Then\[\sum_{n=0}^{\infty}\frac{1+2^n}{3^n}=\sum_{n=0}^{\infty}\left( \frac{1}{3^n}+\left( \frac{2}{3} \right)^n \right)=\sum_{n=0}^{\infty}\frac{1}{3^n}+\sum_{n=0}^{\infty}\left( \frac{2}{3} \right)^n\]You can break the sum up in the last part because the series converges uniformly.
Now you have two geometric series - the first has first term a=1 and common ratio r=1/3, while the second has first term a=1 and common ratio 2/3. The formula for the sum of geometric series then gives for each,\[\frac{1}{1-1/3}+\frac{1}{1-2/3}=\frac{9}{2}\]