<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] Let $S$ be the sum in question.
Let $f(x)=(1+x)^n+(1-x)^n$.
Then $f'(1)=2S$.
Now $f'(x)=n(1+x)^{n-1}-n(1-x)^{n-1}$ and so $f'(1)=n 2^{n-1}$.
Therefore, $S=n 2^{n-2}$.
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