How to prove that $\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n)$ I know that $$\sum_{k=0}^n \binom nk k^2=2^{n-2}(n^2+n),$$ but I cannot find a way how to prove it. I tried induction but it did not work. On wiki they say that I should use differentiation but I do not know how to apply it to binomial coefficient. 
Thanks for any response.
(Presumptive) Source: Theoretical Exercise 1.12(b), P18, A First Course in Pr, 8th Ed, by S Ross

Dilip's comment points to the standard way. Here goes an alternative way:
If you are familiar with basic probability, you should recognize a canonical Binomial distribution in $$ \frac{1}{2^n} {n \choose k}$$
which count the number the successes in $n$ trials of an experiment with prob. $=1/2$ . This Binomial can also be regarded as the sum of $n$ iid Bernoulli variables with $p=1/2$. Hence, the mean is $n p = n/2$ and the variance $n p (1-p)= n/4$
Hence the second moment is 
$$ E[X^2]=\frac{1}{2^n} \sum_{k=0}^n {n \choose k} k^2 = (E[X])^2+\sigma^2_X = n^2 \frac{1}{4} + n \frac{1}{4} = 2^{-2}(n^2+n)$$
