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I am asked to calculate the series $$\sum_{i=0}^\infty \frac{i^2}{6^i}$$ From Wolfram Alpha, I know the answer is $\frac{42}{125}$, but I don't know the steps to get the answer. I was told to use $\frac{1}{ 1-a}$.

Can someone help?

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    $\begingroup$ Yes. Derive once and twice the function $6/(6-x) = \sum_1^\infty x^n/6^n$ (which is defined for $|x| < 6$) and put in $x=1$ in the end. $\endgroup$ – Friedrich Philipp Mar 9 '16 at 0:46
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Here is a useful finite evaluation: $$ 1+r+r^2+\cdots+r^n=\frac{1-r^{n+1}}{1-r}, \quad |r|<1. \tag1 $$ Then by differentiating $(1)$ you get $$ 1+2r+3r^2+\cdots+nr^{n-1}=\frac{1-r^{n+1}}{(1-r)^2}+\frac{-(n+1)r^{n}}{1-r}, \quad |r|<1, \tag2 $$ multiplying $(2)$ by $r$ and differentiating gives $$ 1+4r+9r^2+\cdots+n^2r^{n-1}=\frac{1+r-r^n \left(1+2 n+n^2+r-2 n r-2 n^2 r+n^2 r^2\right)}{(1-r)^3}. \tag3 $$

By making $n \to +\infty$ in $(3)$, using $|r|<1$ and multyplying by $r$ we get

$$ \sum_{n=1}^\infty n^2r^n=\frac{r(1+r)}{(1-r)^3},\quad |r|<1. \tag4 $$

Put $r=\dfrac16$ in $(4)$ to obtain the sought result.

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  • $\begingroup$ Ah. I see. My confusion was with the equation he told us to use. It makes much more sense to use the one for when 'a' (or 'r' in your case) does not equal 1. Thank you very much! $\endgroup$ – user321146 Mar 9 '16 at 1:57
  • $\begingroup$ @user321146 You are welcome. $\endgroup$ – Olivier Oloa Mar 9 '16 at 2:37

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