Simplification of ${0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots}$ Simplify
$$0 \binom{n}{0} + 2 \binom{n}{2} + 4 \binom{n}{4} + 6 \binom{n}{6} + \cdots,$$ where $n \ge 2.$
I think we can write this as the summation $\displaystyle\sum_{i=0}^{n} 2i\binom{n}{2i},$ which simplifies to $\boxed{n\cdot2^{n-2}}.$ Am I on the right track?

A combinatorial proof is in order!
The given expression is the number of ways to choose an even-sized subset of $n$ people, and promote one of them to be the leader. Notice that $n\cdot 2^{n-2}$ is the number of ways to choose the leader first, then choose an odd-sized subset from the remaining $n-1$ people (as the number of odd-sized subsets is the same as the number of even-sized ones). Hence the two sides are equal.
