<s>[INST] Series sum of $\sum_{n = 1}^ {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$) What is the sum of following series $\sum_{n = 1}^  {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$)

*

*$\frac{3}{8}$


*$\frac{3}{10}$


*$\frac{3}{14}$


*$\frac{3}{16}$
My Attempt:
$\sum_{n = 1}^  {\infty} (\frac{1}{2^n3^n} + \frac{1}{2^{n+1}3^n}$) = $\sum_{n = 1}^  {\infty}\frac{1}{2^n 3^n} (1 + \frac{1}{2}$)  = $\frac{3}{2}\sum_{n = 1}^  {\infty}\frac{1}{2^n 3^n}$ = $\frac{3}{2}\sum_{n = 1}^  {\infty}\frac{1}{6^n}$ = $\frac{3}{2}\cdot\frac{1}{1 - {1\over6}}$ = $\frac{3}{2}\cdot \frac{6}{5}$ = $\frac{9}{5}$ which is different from all the above options. Please point out my mistake or solve it different and easiest approach. Thanks.
 [/INST] Your error is that the geometric series starts with $n=1$, not $n=0$, so it should be $\sum_{n=1}^\infty \frac{1}{6^n} = \frac{1}{6} \frac{1}{1-\frac{1}{6}} = \frac{1}{5}$, not $\frac{6}{5}$. Making this correction will give you one of the answers in the multiple choice.
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