How do I prove $2 \binom n2 + 4 \binom n4 + 6 \binom n6 + \cdots = n 2^{n-2}$ How do I prove $2 \binom n2 + 4 \binom n4 + 6 \binom n6 + \cdots = n 2^{n-2}$
I thought about using known $\sum k\binom nk = n2^{n-1} $ and then just dividing it by 2. But is it possible you can help me with writing the full solution, because for the first known theorem you need to derivate $(a+b)^n $and also $a=1, b=x$. But for this one it won't work. maybe I can just prove that known theorem I presented, prove it, and then just divide it by 2. because the known theorem is  $1 \binom n2 + 2 \binom n4 + 3 \binom n6 + \cdots$ (co efficients are divided by two than the presented question)
Edit: I posted a solution.
 A: \begin{align*}
(1+x)^n & = \sum_{k=0}^n \binom{n}{k}x^k\\
\frac{d}{dx}(1+x)^n & = \sum_{k=1}^n k\binom{n}{k}x^{k-1}\\
n(1+x)^{n-1} & = \sum_{k=1}^n k\binom{n}{k}x^{k-1}
\end{align*}
Likewise
\begin{align*}
n(1-x)^{n-1} & = \sum_{k=1}^n (-1)^kk\binom{n}{k}x^{k-1}
\end{align*}
Now add 
$$n(1+x)^{n-1} +n(1-x)^{n-1} =2\sum_{\stackrel{k=1}{k \text{ even}}}^nk\binom{n}{k}x^{k-1}$$
and evaluate at $x=1$.
A: Hint: For an "algebraic" argument, note that $k \binom{n}{k}=n\binom{n-1}{k-1}$.
A: For a combinatorial argument, suppose that you have a pool of $n$ candidates, and you want to choose from that pool a team with an even number of members and appoint one of those members captain. Clearly the lefthand side of your identity counts the number of ways to do this, dividing the count up according to the size of the team. However, we can also do it by first choosing a captain and then choosing an odd number of the remaining $n-1$ candidates to fill out the team. There are $n$ ways to choose the captain and $2^{n-2}$ odd-sized subsets of the remaining $n-1$ candidates, so we can carry out the task in $n2^{n-2}$ ways.
Note, though, that this fails if $n<2$: in that case the lefthand side is $0$.
A: Simple solution:
$(1-x)^n = 1 - \binom n1x + \binom n2 x^2 - \binom n3 x^3+ ...$, now we derivate, and then we enter $ x = 1 $. and then we take all the coefficients with negative signs to $LHS$ and we keep the positive ones in $RHS$. opinions?
