Making sense of a combinatorial answer $\frac {\sum^n_{k= 0}k\binom nk}{2^n}$ is the average size of a subset of $\{1, 2, \ldots, n\}$. 
We add up the sizes of all subsets and divide by the total number of subsets. Why is the $i$th term in the numerator is of the for $k\binom nk$? I mean what's the point of $k$ in front of $\binom nk$?
 A: There are $\binom{n}k$ subsets of $\{1,2,\ldots,n\}$ with cardinality $k$, so they contribute a total of $k\binom{n}k$ to the sum of the cardinalities of all subsets of $\{1,2,\ldots,n\}$. The $k$ is the size of a subset; the $\binom{n}k$ is the number of subsets of that size.
A: The average size of the subsets of an $n$ element set, $S$, is given by $$\frac{1}{2^n}\sum_{A \subset S} card(A).$$
We can break this down as follows:
$$\frac{1}{2^n} \sum_{k=0}^n k \cdot \left( \text{Number of subsets with size } k \right)$$
The number of subsets of $S$ with size $k$ is given by ${n \choose k}$ and so $$\frac{1}{2^n} \sum_{k=0}^n k \cdot {n\choose k}$$ is the average size of subsets of $S$.
We can arrive at a closed form for the sum. In particular, note that $$\sum_{k=0}^n {n \choose k}x^k = (1+x)^n.$$ With a little calculus this gives: $$\sum_{k=0}^n {n\choose k} k x^{k-1} = n (1+x)^{n-1}.$$ Evaluating this at $x=1$ yields: $$\sum_{k=0}^n {n\choose k} k = n 2^{n-1}$$ and so the average size of subsets of $S$ is given by $$\frac{n2^{n-1}}{2^n}=\frac{n}{2}.$$
