Would it be possible to evaluate this sum?
$$\sum_{k=0}^{N/2}k\binom{N+1}{k},$$
where $N$ is even? I know that the sum $$\sum_{k=0}^{N+1}k\binom{N+1}{k}=2^N(N+1)$$ (by http://mathworld.wolfram.com/BinomialSums.html, equation $(21)$), but I can't figure out a way to evaluate this sum from just $k=0$ to $N/2$.
Since for $k$ such that $1 \leq k \leq N + 1$, we have $$k\binom{N + 1}{k} = k \frac{(N + 1)!}{k! (N + 1 - k)!} = \frac{(N + 1)!}{(k - 1)!(N + 1 - k)!} = (N + 1)\binom{N}{k - 1}.$$ If $N$ is even, $\binom{N}{0} = \binom{N}{N}, \binom{N}{1} = \binom{N}{N - 1}, \binom{N}{2} = \binom{N}{N - 2}, \ldots, \binom{N}{N/2 - 1} = \binom{N}{N/2 + 1}$ implies that \begin{align*} & \sum_{k = 1}^{N/2}\binom{N}{k - 1} = \binom{N}{0} + \binom{N}{1} + \cdots + \binom{N}{N/2 - 1} \\ = &\frac{\binom{N}{0} + \cdots + \binom{N}{N} - \binom{N}{N/2} }{2} \\ = & \frac{1}{2}\sum_{k = 0}^N \binom{N}{k} - \frac{1}{2}\binom{N}{N/2} \\ = & 2^{N - 1} - \frac{1}{2}\binom{N}{N/2}. \end{align*} Hence $$\sum_{k = 0}^{N/2} k\binom{N + 1}{k} = (N + 1)2^{N - 1} - \frac{1}{2}(N + 1)\binom{N}{N/2}.$$