Given that $n$ is even, find a closed-form expression for $\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k}$ Given that $n$ is even, find a closed-form expression for
$$\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k},$$
or, in other words, for the sum
$$2^2\cdot 2\cdot\binom n2 + 2^4\cdot 4\cdot\binom n4 + 2^6\cdot 6\cdot\binom n6 + \cdots + 2^n\cdot n\cdot\binom nn.$$
You may find the identity $k\binom{n}{k} = n\binom{n-1}{k-1}$ helpful.

I am stuck on this problem!  I tried using the identity but it only confuses me more.  All solutions are greatly appreciated!  Thanks!
 A: By your identity,we have 
$\sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k}=\sum_{k=0}^{n/2}2^{2k}\cdot n\cdot\binom{n-1}{2k-1}=2n\sum_{k=0}^{n/2}2^{2k-1}\cdot\binom{n-1}{2k-1}$
Now 
\begin{equation}
-\sum_{k=0}^{n/2}2^{2k-1}\cdot\binom{n-1}{2k-1}=
\sum_{k=0}^{n/2}(-2)^{2k-1}\cdot\binom{n-1}{2k-1}=\frac{1}{2}\big((-2+1)^{n-1}-(2+1)^{n-1}\big)
\end{equation}
you can check the second '=' using the binomial expansion(basically the even terms cancel due to the minus sign before 2). Hence your expression is $n\cdot (3^{n-1}+1)$.
A: $$
S = \sum_{k=0}^{n/2} 2^{2k}\cdot 2k\cdot \binom{n}{2k} = 2 n \sum_{k=1}^{n/2} 2^{2k-1}\cdot \binom{n-1}{2k-1}
$$
We have by the Binomial theorem:
$$
B =  2n \sum_{k=0}^{n-1} 2^{k}\cdot \binom{n-1}{k} = 2n \cdot 3^{n-1}
$$
and also (n even)
$$
 R = 2n \sum_{k=0}^{n-1} (-1)^{n-1-k}2^{k}\cdot \binom{n-1}{k} \\=  2n [ - 2^0 \cdot \binom{n-1} {0} +  2 \cdot \binom{n-1} {1} - 2^2 \cdot \binom{n-1} {2} + \cdots]=  2n 
$$
Adding the two  gives $B+R = 2S$, hence 
$S = n (3^{n-1} +1)$
