Induction on binomial Identity: $0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$ I am having trouble proving the following identity:
$0\cdot {n\choose 0} + 2\cdot {n\choose 2} + 4\cdot {n\choose4}+\ldots = n\cdot2^{n-2}$
Here is what I have so far:
Proof:
Base: Let $n=0$:
LHS: $0\cdot {0\choose 0} = 0\cdot1 = 0$
RHS: $0\cdot 2^{0-2} = 0$
Step: Let $k\in \mathbb{Z} s.t k \geq 0$ and assume the identity is true for k.
Consider the LHS for $k+1$ where $k$ is even (I leave out the odd case because I think it will turn out the same?):
\begin{align}
=& 0\cdot{k+1\choose 0}+2\cdot{k+1\choose 2}+4\cdot{k+1\choose 4}+... +k\cdot{k+1\choose k}
\\=&0\cdot\left[{k\choose 0}+{k\choose -1}\right] + 2\cdot\left[{k\choose 2}+{k\choose 1}\right]+ 4\cdot\left[{k\choose 4}+{k\choose 3}\right]+\ldots+ k\cdot\left[{k\choose k}+{k\choose k-1}\right]
\\=&\left[0\cdot{k\choose 0}+2\cdot {k\choose 2}+4\cdot{k\choose 4}+\ldots+k\cdot{k\choose k}\right] + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\=& k\cdot2^{k-2} + \left[0\cdot{k\choose -1}+ 2\cdot {k\choose 1}+4\cdot {k\choose 3}+\ldots+k\cdot{k\choose k-1}\right]
\\
\end{align}
I know I need to end up with something like:
\begin{align}
=&k\cdot2^{k-2}+ \left[k\cdot2^{k-2} + 2^{k-1}\right]
\\=&2k\cdot 2^{k-2} + 2^{k-1}
\\=&k\cdot 2^{k-1}+2^{k-1}
\\=&(k+1)\cdot 2^{k-1}\end{align} 
But, how can I get what I need from the combinations above? It may not end up exactly like that, but what is the reasoning behind this? 

Add the first two to get the third:
$$
\begin{align}
\sum_{k=0}^nk\binom{n}{k}&=\sum_{k=0}^nn\binom{n-1}{k-1}&=n(1+1)^{n-1}&=n2^{n-1}\\
\sum_{k=0}^n(-1)^kk\binom{n}{k}&=\sum_{k=0}^n(-1)^kn\binom{n-1}{k-1}&=n(1-1)^{n-1}&=0\\
\sum_{k=0}^{\lfloor n/2\rfloor}2k\binom{n}{2k}&&&=n2^{n-1}\\
\end{align}
$$
