<s>[INST] 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?
 [/INST] Spoilered answer:

 $$\sum_{i} 2i \binom{n}{2i}=n\sum_{i} \binom{n-1}{2i-1}= n2^{n-2}$$

To get the second equaltiy note that $\sum_{i}\binom{n-1}{2i-1}=\sum_{i} \binom{n-1}{2i}$ as results from writing the binomial formula for $(1-1)^{n-1}$, and recall that
$$\sum_{i}\binom{n-1}{2i-1}+\sum_{i} \binom{n-1}{2i}= \sum_{i} \binom{n-1}{i}= 2^{n-1}$$
as follows from using binomial formula for $(1+1)^{n-1}$.
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