<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] Here's an alternative approach, motivated by the appearance of even numbers in the summand.
Because $$\frac{1+(-1)^k}{2}=\begin{cases}1&\text{if $k$ is even}\\0&\text{if $k$ is odd}\end{cases}$$
we have $$\sum_{k\ge 0} a_{2k} = \sum_{k\ge 0} a_k \frac{1+(-1)^k}{2}.$$
Now take $a_k=k\binom{n}{k}$ to obtain
\begin{align}
\sum_{k\ge 0} 2k\binom{n}{2k} 
&=\sum_{k\ge 0} k\binom{n}{k}\frac{1+(-1)^k}{2} \\
&=\sum_{k\ge 1} n\binom{n-1}{k-1}\frac{1+(-1)^k}{2} \\
&=\frac{n}{2}\sum_{k\ge 1} \binom{n-1}{k-1} +\frac{n}{2}\sum_{k\ge 1} \binom{n-1}{k-1}(-1)^k \\
&=\frac{n}{2}2^{n-1} +\frac{n}{2}(1-1)^{n-1} \\
&=\frac{n}{2}2^{n-1} +\frac{n}{2}[n=1] \\
&=n2^{n-2}
\end{align}
for $n \ge 2$.
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