we know Bernoulli number such identity $$\sum_{k=0}^{n}\binom{n+1}{k}B_{k}=0$$ see:Bernoulli number identity
show that $$\sum_{k=0}^{2m}\binom{2m+1}{k}\cdot 2^k\cdot B_{k}=0$$
where $B_{n}$ is Bernoulli number.
My idea:since $$B_{0}+\binom{n+1}{1}B_{1}+\binom{n+1}{2}B_{2}+\cdots+\binom{n+1}{n}B_{n}=0$$ and $$\Longleftrightarrow \sum_{k=0}^{2m}\binom{2m+1}{k}\cdot 2^k\cdot B_{k}=\sum_{k=0}^{2m}\dfrac{(2m+1)!}{k!(2m+1-k)!}\cdot 2^k\cdot B_{k}=0$$
then I can't Continue.Thank you
The sequence $(B_n)_n$ has a generating function: $$ \sum_{n=0}^ \infty \frac{B_n}{n!}x^n=\frac{x}{e^x-1}~\buildrel{\rm def}\over{=}~G(x) $$ see here. Hence, $$ e^xG(2x)=\frac{x}{2\sinh x} $$ and this is an even function. So, the coefficient of $x^{2m+1}$ in the power series expansion of $e^xG(2x)$ is $0$. That is $$ \sum_{k=0}^{2m+1}\binom{2m+1}{k}2^kB_k=0 $$ Finally, it is well-known that $B_{2m+1}=0$ for $m\geq 1$, hence $$ \forall\,m\geq 1,\qquad\sum_{k=0}^{2m}\binom{2m+1}{k}2^kB_k=0 $$ Of course the result is not correct when $m=0$.