How to prove this Bernoulli summation

Question

I have tried to use the techniques from question 822466 and 135844 but I don't get it right.

For all $d\in\mathbb{N}$ prove

$$\sum_{j=0}^{2d+1} \binom{2d+1}{j}\frac{B_{j+1}}{j+1}(4^{j+1}-2^{j+1})=0$$

How can this be done?

Answer 1

The given sum equals $$ (2d+1)!\sum_{j=0}^{2d+1}\frac{B_{j+1}}{(j+1)!}(4^{j+1}-2^{j+1})\frac{1}{(2d+1-j)!}. \tag{1}$$ We have $$ \frac{1}{e^x-1}-\frac{1}{x} = \sum_{j\geq 0}\frac{B_{j+1}}{(j+1)!} x^j \tag{2}$$ hence $(1)$ can be written as $(2d+1)!$ times the coeffient of $x^{2d+1}$ in $$ e^x\left[\frac{4}{e^{4x}-1}-\frac{4}{4x}-\frac{2}{e^{2x}-1}+\frac{2}{2x}\right] =-\frac{1}{\cosh(x)}\tag{3}$$ Since $2d+1$ is odd and $-\frac{1}{\cosh(x)}$ is an even function, the claim is trivial after $(3)$.
The generalized sum (with $2d$ in place of $2d+1$) is related with Euler numbers.