14. [40] Find an explicit, closed form formula for
$$
\sum_{k=1}^{n} \frac{k \cdot(-1)^{k} \cdot\binom{n}{k}}{n+k+1}
$$
Answer: $\frac{-\mathbf{1}}{\binom{2 \mathrm{n}+1}{\mathbf{n}}}$ or $-\frac{\mathbf{n}!(\mathbf{n}+\mathbf{1})!}{(2 \mathbf{n}+1)!}$ or obvious equivalent.
Solution. Consider the interpolation of the polynomial $P(x)=x \cdot n$ ! at $x=0,1, \ldots, n$. We obtain the identity
$$
\begin{aligned}
P(x) & =x \cdot n!=\sum_{k=0}^{n} k \cdot n!\prod_{j \neq k} \frac{x-j}{k-j} \\
& =\sum_{k=0}^{n} k \cdot n!\cdot \frac{x(x-1) \cdots(x-k+1)(x-k-1) \cdots(x-n)}{k!(n-k)!(-1)^{n-k}} \\
& =\sum_{k=1}^{n} k \cdot(-1)^{n-k} \cdot\binom{n}{k} \cdot x(x-1) \cdots(x-k+1)(x-k-1) \cdots(x-n) .
\end{aligned}
$$

This identity is valid for all complex numbers $x$, but, to extract a factor $\frac{1}{n+k+1}$ from the valid product of each summand, we set $x=-n-1$, so that
$$
-(n+1)!=\sum_{k=1}^{n} k(-1)^{n-k}\binom{n}{k}(-n-1) \cdots(-n-k)(-n-k-2) \cdots(-2 n-1)=\sum_{k=1}^{n} \frac{k(-1)^{k}\binom{n}{k}(2 n+1)!}{n!(n+k+1)} \text {. }
$$

Finally,
$$
\sum_{k=1}^{n} \frac{k \cdot(-1)^{k} \cdot\binom{n}{k}}{n+k+1}=\frac{-n!(n+1)!}{(2 n+1)!}=\frac{-1}{\binom{2 n+1}{n}}
$$