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.

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    $\begingroup$ All $B_{2k+1}$, except $B_1$ are $0$. So you get $B_n - 2nB_1$. $\endgroup$ – Daniel Fischer Nov 5 '16 at 17:51

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

$$\bbox[5px,border:2px solid #00A000]{\frac{n!}{k!} [z^{n+1-k}] \frac{z \exp(z)}{1-\exp(-z)}.}$$


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