An infinite summation problem

First note that for l $x$ l < 1 we have

$\sum_{k=1}^\infty$ ( $x^{k}$ ) = $x / (1 - x)$ .

Now differentiate both sides and multiply through by $x$ to find that

$\sum_{k=1}^\infty$ ( $k * x^{k}$ ) = x / $(1 - x)^2$ .

Differentiate both sides again and multiply through by $x$ to find that

$\sum_{k=1}^\infty$ ( $k^{2} * x^{k}$ ) = $x * (1 + x)$ / $(1 - x)^{3}$ .

Now just plug in $x = 3/4$ to find that the sum comes to $\boxed{84}$ .

I realize that this is more of a calculus approach as opposed to algebraic, but it was the most efficient method I could think of applying to the problem.

$\begin{aligned} S & = \sum_{\color{#3D99F6}k=1}^\infty k^2 \left(\frac 34 \right)^k \\ & = \sum_{\color{#D61F06}k=0}^\infty (k+1)^2 \left(\frac 34 \right)^{k+1} \\ & = \frac 34 {\color{#3D99F6} \sum_{\color{#D61F06}k=0}^\infty k^2 \left(\frac 34 \right)^k} + \frac 32 \sum_{\color{#D61F06}k=0}^\infty k \left(\frac 34 \right)^k + \frac 34 \sum_{\color{#D61F06}k=0}^\infty k \left(\frac 34 \right)^k & \small \color{#3D99F6} \text{Note that } S = \sum_{k=1}^\infty k^2 \left(\frac 34 \right)^k = \sum_{\color{#D61F06}k=0}^\infty k^2 \left(\frac 34 \right)^k \\ & = \frac 34 {\color{#3D99F6} S} + \frac 32 \sum_{\color{#D61F06}k=0}^\infty k \left(\frac 34 \right)^k + \frac 34 \left(\frac 1{1-\frac 34} \right) \end{aligned}$

$\begin{aligned} \implies \frac S4 & = \frac 32 {\color{#3D99F6} \sum_{k=0}^\infty k \left(\frac 34 \right)^k} +3 & \small \color{#3D99F6} \text{Using the same method (see note).} \\ & = \frac 32 {\color{#3D99F6} (12)} +3 \\ \implies S & = \boxed{84} \end{aligned}$

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Note:

$\small \begin{aligned} S_1 & = \sum_{\color{#D61F06} k=0}^\infty k \left(\frac 34 \right)^k = \sum_{\color{#3D99F6}k=1}^\infty k \left(\frac 34 \right)^k \\ & = \sum_{\color{#D61F06} k=0}^\infty (k+1) \left(\frac 34 \right)^{k+1} \\ & = \frac 34 \sum_{\color{#D61F06} k=0}^\infty k\left(\frac 34 \right)^k + \frac 34 \sum_{\color{#D61F06} k=0}^\infty \left(\frac 34 \right)^k \\ & = \frac 34 S_1 + 3 \\ \implies \frac {S_1}4 & = 3 \\ \implies S_1 & = 12 \end{aligned}$