What is the sum $\sum_{k=0}^{n}k^2\binom{n}{k}$? What should be the strategy to find
$$\sum_{k=0}^{n}k^2\binom{n}{k}$$
Can this be done by making a series of $x$ and integrating?

Let $$\displaystyle S=\sum^{n}_{k=0}k^2 \binom{n}{k} \;,$$ Now using $\displaystyle \bullet \; \binom{n}{k} = \frac{n}{k}\cdot \binom{n-1}{k-1}$
So we get $$\displaystyle S = \sum^{n}_{k=0}k^2 \cdot \frac{n}{k}\binom{n-1}{k-1}\;,$$ Again using $\displaystyle \bullet \; \binom{n-1}{k-1} = \frac{n-1}{k-1}\cdot \binom{n-2}{k-2}$
So $$\displaystyle S = n\sum^{n}_{k=1}k\binom{n-1}{k-1} = n\sum^{n}_{k=0}[(k-1)+1]\binom{n-1}{k-1}$$
So we get $$\displaystyle S=n\sum^{n}_{k=1}(k-1)\binom{n-1}{k-1}+n\sum^{n}_{k=1} \binom{n-1}{k-1}$$
So we get $$\displaystyle S=n\sum^{n}_{k=2}(k-1)\cdot \frac{n-1}{k-1}\cdot \binom{n-2}{k-2}+n\sum^{n}_{k=1}\binom{n-1}{k-1}$$ 
So we get $$\displaystyle S=n(n-1)\sum^{n}_{k=0}\binom{n-2}{k-2}+n\sum^{n}_{k=0}\binom{n-1}{k-1}$$
Now Using $\displaystyle \bullet\; \sum^{n}_{k=0}\binom{n}{k} = 2^{k}$
So we get $$\displaystyle S = n(n-1)\cdot 2^{n-2}+n\cdot 2^{n-1}$$
