An identity for Bernoulli numbers Denote by $B_n$ the Bernoulli sequence (defined by the exponential generating function $\frac{x}{e^x-1}=\sum_{n=0}^{\infty}\frac{B_n}{n!}x^n$). As we know
$$
\sum^{n}_{j=0}\binom{n}{j}B_j=B_n\; ; \; n\geq 2
$$
what about $\sum^{n}_{j=0}(-1)^j\binom{n}{j}B_j$, and
as a more general case 
$$
\sum^{n}_{j=0}\frac{(-1)^{j+1-k}}{j+1}\binom{n}{j}\binom{j+1}{k}B_{j+1-k}
$$
where $k$ is a given integer such that $1\leq k\leq n+1$
(is there any similar identity?).
Note that putting $k=1$ we get the first summation.
 A: Suppose we have the Bernoulli numbers defined by
$$\frac{z}{\exp(z)-1} = \sum_{n\ge 0} B_n \frac{z^n}{n!}$$
and ask about a closed form for the quantity
$$\sum_{j=0}^n \frac{(-1)^{j+1-k}}{j+1} 
{n\choose j} {j+1\choose k} B_{j+1-k} 
\\ = \sum_{j=k-1}^n \frac{(-1)^{j+1-k}}{j+1} 
{n\choose j} {j+1\choose k} B_{j+1-k}.$$
This is
$$\frac{1}{k} \sum_{j=k-1}^n (-1)^{j+1-k}
{n\choose j} {j\choose k-1} B_{j+1-k}.$$
Now we have
$${n\choose j} {j\choose k-1}
= \frac{n!}{(n-j)! (k-1)! (j+1-k)!}
= {n\choose k-1} {n+1-k\choose n-j}$$
so we get
$$\frac{1}{k} {n\choose k-1}
\sum_{j=k-1}^n (-1)^{j+1-k}
{n+1-k\choose n-j} B_{j+1-k}
\\ = \frac{1}{k} {n\choose k-1}
\sum_{j=0}^{n+1-k} (-1)^{j}
{n+1-k\choose n+1-k-j} B_{j}
\\ = \frac{1}{k} {n\choose k-1}
\sum_{j=0}^{n+1-k} (-1)^{j}
{n+1-k\choose j} B_{j}.$$
We thus obtain the closed form
$$\frac{1}{k} {n\choose k-1} 
\times (n+1-k)! [z^{n+1-k}] \exp(z) \frac{z}{1-\exp(-z)}.$$
This is
$$\bbox[5px,border:2px solid #00A000]{\frac{n!}{k!} 
[z^{n+1-k}] \frac{z \exp(z)}{1-\exp(-z)}.}$$
As noted in the comments this will simplify to
$$[[k=n]] + [[k\le n]] \times {n\choose k} 
+ {n+1\choose k} \frac{B_{n+1-k}}{n+1}.$$
