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? 
 A: A non-inductive proof now. We have:
$$
\sum_{0 \le k \le n} \binom{n}{k} z^k = (1 + z)^n
$$
So we also have:
$$
\sum_{0 \le k \le \lfloor n / 2 \rfloor} \binom{n}{2 k} z^{2 k} = \frac{(1 + z)^n + (1 - z)^n}{2}
$$
If you differentiate this with respect to $z$ you get:
$$
\sum_{0 \le k \le \lfloor n / 2 \rfloor} 2 k \binom{n}{2 k} z^{2 k - 1}
  = \frac{n (1 + z)^{n - 1} - n (1 - z)^{n - 1}}{2}
$$
Then evaluate at $z = 1$ you get the requested sum:
$$
\sum_{0 \le k \le \lfloor n / 2 \rfloor} 2 k \binom{n}{2 k}
 = \frac{n 2^{n - 1}}{2} = n 2^{n - 2}
$$
A: Here is a proof, which relies on induction indirectly.
Note that 
$$2k \dbinom{n}{2k} = 2k \dfrac{n!}{(n-2k)!(2k)!} = n \dfrac{(n-1)!}{(n-2k)!(2k-1!)} = n \dbinom{n-1}{2k-1}$$
Hence,
$$\sum_{k=1}^{\lfloor n/2 \rfloor} 2k \dbinom{n}{2k} = n\sum_{k=1}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1}$$
Now note that
$$(1+1)^{n-1} = \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k} + \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} \,\,\,\,\, (\heartsuit)$$
and
$$(1-1)^{n-1} = \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k} - \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} \,\,\,\,\, (\spadesuit)$$
Hence,
$$(\heartsuit) - (\spadesuit) \implies 2 \sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} = 2^{n-1}$$
Hence, $$\sum_{k=0}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} = 2^{n-2}$$
Therefore, we get
$$\sum_{k=1}^{\lfloor n/2 \rfloor} 2k \dbinom{n}{2k} = n\sum_{k=1}^{\lfloor n/2 \rfloor} \dbinom{n-1}{2k-1} = n2^{n-2}$$
A: Unlike with other proof methods, proofs by induction sometimes become easier if you make the statement to be proved stronger. In the present case, it seems that your approach should be successful if you add a corresponding statement about the sum of binomial coefficients with odd lower arguments to the claim:
$$
\sum_{j=1}^{\lceil n/2\rceil}2j\binom n{2j-1}=2^{n-2}(n+2)\;.
$$
A: Add the first two to get the third:
$$
\begin{align}
\sum_{k=0}^nk\binom{n}{k}&=\sum_{k=0}^nn\binom{n-1}{k-1}&=n(1+1)^{n-1}&=n2^{n-1}\\
\sum_{k=0}^n(-1)^kk\binom{n}{k}&=\sum_{k=0}^n(-1)^kn\binom{n-1}{k-1}&=n(1-1)^{n-1}&=0\\
\sum_{k=0}^{\lfloor n/2\rfloor}2k\binom{n}{2k}&&&=n2^{n-1}\\
\end{align}
$$
