TITLE: Seeking a combinatorial proof $\sum\limits _{k=0}^n (n-2k)^2\binom{n}{k}=n\times 2^n$
QUESTION [5 upvotes]: I would appreciate if somebody could help me with the following problem
Q: Seeking a combinatorial proof $(\binom{n}{k}=\frac{n!}{k! (n-k)!}
)$
$$\sum _{k=0}^n (n-2 k)^2 \binom{n}{k}=n\times
 2^n$$

REPLY [5 votes]: Consider all subsets of a set of $n$ elements, along with their complements; count $+1$ for each ordered pair of (possibly identical) elements that are in the same set (i.e. both in the subset or both in its complement) and $-1$ for each ordered pair of elements that aren't. Then for a subset of size $k$, of which there are $\binom nk$, the sum is $((n-k) - (k))^2=(n-2k)^2$. A pair of different elements doesn't contribute to the sum over all subsets, since they are in the same set just as often as they are in different sets. A pair of identical elements contributes $+1$ for each of the $2^n$ subsets, and there are $n$ such pairs.