By sum properties, prove that: $$\sum_{k=1}^nk^2\, 2^{-k}=2^{-n}(-6+3\cdot 2^{1+n}-4n-n^2)$$
Progress so far: $$\sum_{k=1}^nk^2\,2^{-k} = 1\cdot (1/2) + 4\cdot (1/4) + 9\cdot (1/8) + 16\cdot (1/16) + 25\cdot (1/32) + 36\cdot (1/64)+\dots +n^2\cdot (1/2^n)$$
Let $S(n) = \sum_{k=1}^n2^{-k}k^2$. $$ \begin{align*} S(n+1) &= S(n)+2^{-n-1}(n+1)^2\\ &= \sum_{k=1}^{n+1}2^{-k}k^2\\ &= \frac{1}{2} + \sum_{k=2}^{n+1}2^{-k}k^2\\ &= \frac{1}{2} + \sum_{k=1}^{n}2^{-k-1}(k+1)^2\\ &= \frac{1}{2} + \frac{1}{2}\sum_{k=1}^n2^{-k}(k+1)^2\\ &= \frac{1}{2} + \frac{1}{2}\sum_{k=1}^n2^{-k}(k^2 + 2k + 1)\\ &= \frac{1}{2} + \frac{1}{2}\left(\sum_{k=1}^n2^{-k}k^2 + \sum_{k=1}^n2^{-k}2k + \sum_{k=1}^n2^{-k}\right)\\ &= \frac{1}{2} + \frac{1}{2}\left(S(n) + 2\sum_{k=1}^n2^{-k}k + \sum_{k=1}^n2^{-k}\right) \end{align*} $$ Now use the same technique to find $A(n) = \sum_{k=1}^n2^{-k}k$ and $\sum_{k=1}^n2^{-k}$ is a geometric series. Then finally solve the above equation for $S(n)$. $$ \begin{align*} A(n+1) &= A(n) + 2^{-n-1}(n+1)\\ &=\sum_{k=1}^{n+1} 2^{-k}k\\ &= \frac{1}{2}+ \sum_{k=2}^{n+1}2^{-k}k\\ &= \frac{1}{2}+ \sum_{k=1}^{n}2^{-k-1}(k+1)\\ &= \frac{1}{2}+ \frac{1}{2}\sum_{k=1}^{n}2^{-k}(k+1)\\ &= \frac{1}{2}+ \frac{1}{2}\left(\sum_{k=1}^{n}2^{-k}k + \sum_{k=1}^{n}2^{-k}\right)\\ &= \frac{1}{2}+ \frac{1}{2}\left(A(n) + \sum_{k=1}^{n}2^{-k}\right) \end{align*} $$ So now solve for $A(n)$ and finally $S(n)$.