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?

Sure. Notice that
$$\sum _{i = 0}^{n}2i\binom{n}{2i}=\sum _{i = 0}^{n}\binom{2i}{1}\binom{n}{2i}=\sum _{i = 1}^{n}\binom{n}{1}\binom{n-1}{2i-1}=n\sum _{i=1}^n\binom{n-1}{2i-1}=n\cdot 2^{n-2}.$$
Notice that the last step is because you are adding half of the binomials, and the odd half equals the even half by the binomial theorem.
