#19 Hua Luo Geng.

$\displaystyle\sum_{n=0}^{\infty}x^n=\dfrac{1}{1-x}\phantom{cccccc} |x|<1$

Taking the derivative on both sides and multiplying by $x$

$\displaystyle\sum_{n=0}^{\infty}nx^{n-1}=\dfrac{1}{(1-x)^2}$

$\displaystyle\sum_{n=0}^{\infty}nx^n=\dfrac{x}{(1-x)^2}$

Taking the derivative and multiplying by $x$ again

$\displaystyle\sum_{n=0}^{\infty}n^2x^{n-1}=\dfrac{x+1}{(1-x)^3}$

$\displaystyle\sum_{n=0}^{\infty}n^2x^n=\dfrac{x(x+1)}{(1-x)^3}$

We now have

$S(x)=\displaystyle\sum_{n=0}^{\infty}n^2x^n=\dfrac{x(x+1)}{(1-x)^3}$

We have to find

$S\left(\dfrac{1}{2}\right)=\dfrac{\dfrac{1}{2}\left(\dfrac{1}{2}+1\right)}{\left(1-\dfrac{1}{2}\right)^3}=\boxed{6}$

Let $\displaystyle \sum_{n \ = \ 1}^{\infty} \frac {n^2}{2^n} = S$ .

Note that

$\displaystyle S = \sum_{n \ = \ 1}^{\infty} \frac {n^2}{2^n} = \sum_{n \ = \ 0}^{\infty} \frac {n^2}{2^n}$

and

$\displaystyle S = \sum_{n \ = \ 1}^{\infty} \frac {n^2}{2^n} = \sum_{n \ = \ 0}^{\infty} \frac {(n + 1)^2}{2^{n + 1}}$

Thus,

$\displaystyle S = \sum_{n \ = \ 0}^{\infty} \frac {n^2}{2^n} = \sum_{n \ = \ 0}^{\infty} \frac {(n + 1)^2}{2^{n + 1}}$

Now let's do some rewriting:

$\displaystyle \begin{aligned} S & = 2S - S \\ & = \sum_{n \ = \ 0}^{\infty} \frac {(n + 1)^2}{2^n} - \sum_{n \ = \ 0}^{\infty} \frac {n^2}{2^n} \\ & = \sum_{n \ = \ 0}^{\infty} \frac {2n + 1}{2^n} \\ & = \sum_{n \ = \ 0}^{\infty} \frac {2n}{2^n} + \sum_{n \ = \ 0}^{\infty} \frac {1}{2^n} \\ & = 2\sum_{n \ = \ 0}^{\infty} \frac {1}{2^n} + \sum_{n \ = \ 0}^{\infty} \frac {1}{2^n} \\ & = 4 + 2 \\ & = \boxed{6} \end{aligned}$

$\begin{aligned} S & = \sum_{\color{#3D99F6}n=1}^\infty \frac {n^2}{2^n} = \sum_{\color{#D61F06}n=0}^\infty \frac {n^2}{2^n} \\ & = \sum_{\color{#D61F06} n=0}^\infty \frac {(n+1)^2}{2^(n+1)} \\ & = \sum_{\color{#D61F06} n=0}^\infty \frac {n^2+2n+1}{2\cdot 2^n} \\ & = \frac 12 \sum_{\color{#D61F06} n=0}^\infty \frac {n^2}{2^n} + \frac 12{\color{#3D99F6}\sum_{n=1}^\infty \frac n{2^{n-1}}} + \frac 12 \sum_{\color{#D61F06} n=0}^\infty \frac 1{2^n} & \small \color{#3D99F6} \text{Note that } \sum_{n=1}^\infty \frac n{2^{n-1}} \text{ is sum of AGP.} \\ & = \frac 12 S + \frac 12 \times {\color{#3D99F6}4} + \frac 12 \left(\frac 1{1-\frac 12}\right) & \small \color{#3D99F6} \text{See note.} \\ & = \frac 12 S + 2 + 1 \\ \implies \frac 12 S & = 3 \\ S & = \boxed{6} \end{aligned}$

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Note: Arithmetic-geometric progression (AGP)

$\displaystyle \sum_{k=1}^n [a+(k-1)d]r^{k-1} = \frac {a-[a+(n-1)d]r^n}{1-r} + \frac {dr(1-r^{n-1})}{(1-r)^2}$