Ah, it's k^2/2^k

$\begin{aligned} \mathbb S&=\sum_{k=1}^\infty\frac{k^2}{2^k}\\ &=\frac 12\color{#3D99F6}\sum_{k=1}^\infty\frac{(k-1)^2}{2^{k-1}}\color{#333333}+\sum_{k=1}^\infty\frac{(2k-1)}{2^k}\\ &=\frac 12\color{#3D99F6}\mathbb S\color{#333333}+\color{#D61F06}\sum_{k=1}^\infty\frac{k}{2^k}\color{#333333}+\frac 12\color{#D61F06}\sum_{k=1}^\infty\left(\frac{k-1}{2^{k-1}}\right)\\ &=\frac 12\color{#3D99F6}\mathbb S\color{#333333}+\frac 32\color{#D61F06}\sum_{k=1}^\infty\frac{k}{2^k}\\ \mathbb S-\frac 12\color{#3D99F6}\mathbb S&\color{#333333}=\frac 32\color{#D61F06}\sum_{k=1}^\infty\frac{k}{2^k}\\ \end{aligned}$ So we can write $\begin{aligned} \frac 13\color{#3D99F6}\mathbb S&\color{#333333}=\color{#D61F06}\sum_{k=1}^\infty\frac{k}{2^k}\\ &=\frac 12\color{#D61F06}\sum_{k=1}^\infty\frac{k-1}{2^{k-1}}\color{#333333}+\color{#20A900}\sum_{k=1}^\infty\frac{1}{2^k}\\ &=\frac 12\left(\frac 13\color{#3D99F6}\mathbb S\right)\color{#333333}+\color{#20A900}1\\ \mathbb S&=6 \end{aligned}$

Start with

$\begin{array}{l} \sum_{k=0}^{\infty} x^k = \frac{1}{1-x} \\ { \color{#3D99F6} \frac{d}{dx} } \sum_{k=0}^{\infty} x^k ={ \color{#3D99F6} \frac{d}{dx} } \frac{1}{1-x} \\ { \color{#3D99F6} x } \sum_{k=1}^{\infty} kx^{k-1} ={ \color{#3D99F6} x } \frac{1}{(1-x)^2} \\ { \color{#3D99F6} \frac{d}{dx} } \sum_{k=1}^{\infty} kx^{k} ={ \color{#3D99F6} \frac{d}{dx} }\frac{x}{(1-x)^2} \\ { \color{#3D99F6} x } \sum_{k=1}^{\infty} k^2x^{k-1} = { \color{#3D99F6} x } \frac{1+x}{(1-x)^3} \\ \sum_{k=1}^{\infty} k^2x^{k} = \frac{x(1+x)}{(1-x)^3} \\ \end{array}$

Therefore,

$\begin{array}{l} \sum_{k=1}^{\infty} \frac{k^2}{2^k} = \sum_{k=1}^{\infty} k^2(\frac{1}{2})^k \\ = \frac{(\frac{1}{2})(1+\frac{1}{2})}{(1-\frac{1}{2})^2} = 6 \\ \end{array}$

$\begin{aligned} S & = \sum_{\color{#3D99F6}k=1}^\infty \frac {k^2}{2^k} = \sum_{\color{#D61F06}k=0}^\infty \frac {k^2}{2^k} \\ & = \sum_{\color{#D61F06}k=0}^\infty \frac {(k+1)^2}{2^{k+1}} \\ & = \frac 12 \sum_{k=0}^\infty \frac {k^2+2k+1}{2^k} \\ & = \frac 12 \sum_{k=0}^\infty \frac {k^2}{2^k} + {\color{#3D99F6}\sum_{\color{#D61F06}k=0}^\infty \frac k{2^k}} + \frac 12 \sum_{k=0}^\infty \frac 1{2^k} & \small \color{#3D99F6} \text{Again }\sum_{\color{#D61F06}k=0}^\infty \frac k{2^k} = \sum_{k=1}^\infty \frac k{2^k} = S_1 \\ & = \frac S2 + {\color{#3D99F6}\sum_{k=0}^\infty \frac {k+1}{2^{k+1}}} + \frac 12 \left(\frac 1{1-\frac 12}\right) \\ & = \frac S2 + {\color{#3D99F6}\frac 12 \sum_{k=0}^\infty \frac k{2^k} + \frac 12 \sum_{k=0}^\infty \frac 1{2^k}} + 1 \\ & = \frac S2 + {\color{#3D99F6}\underbrace{\frac {S_1}2 + 1}_{=S_1}} + 1 & \small \color{#3D99F6} \text{Since } S_1 = \frac {S_1}2+1 \implies S_1 = 2 \\ & = \frac S2 + {\color{#3D99F6}2} + 1 \\ & = \boxed{6} \end{aligned}$