How to prove that:
$$\sum_{k=0}^n \binom n k 2^k B_k = (2-2^n)B_n$$
In this sum, $B_n$ is the Bernoulli number with $B_1 = -\frac 1 2$. Thanks for your attention!
Sum with Bernoulli numbers
bernoulli numberscombinatoricsgenerating-functionssequences-and-series
Best Answer
The identity that you want to verify can be rewritten as $$\sum^n_{k=0}\frac{2^k B_k}{k!(n-k)!} = 2\frac{B_n}{n!} - \frac{2^nB_n}{n!}$$ From the (standard) definition of Bernoulli numbers $B_n$, $g(z):=\frac{z}{e^z-1}=\sum^\infty_{n=0}\frac{B_n}{n!}z^n$. Hence, the right hand side of the first equation above corresponds to $n$-th coefficient of the power series \begin{align} f(z)&=2g(z) - g(2z)\\ &=2\frac{z}{e^z-1}-\frac{2z}{e^{2z}-1}\\ &=\frac{2z}{e^{2z} -1}e^z = g(2z)e^z\\ &=\Big(\sum^\infty_{n=0}\frac{B_n}{n!}2^nz^n\Big)\Big(\sum^\infty_{n=0}\frac{1}{n!}z^n\Big)\\ &=\sum^\infty_{n=0}c_nz^n \end{align} where $c_n=\sum^n_{k=0}\frac{2^kB_k}{k!}\frac{1}{(n-k)!}$ which is the left hands side of the identity we are trying to verify.