Sum with Bernoulli numbers

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.

One possible way to demonstrate the thesis is through the B. Polynomials $$ \eqalign{ & S(n) = \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr k \cr} \right)2^{\,k} B_{\,k} } = 2^{\,n} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr n - k \cr} \right)\left( {{1 \over 2}} \right)^{\,n - k} B_{\,k} } = \cr & = 2^{\,n} \sum\limits_{\left( {0\, \le } \right)\,k\,\left( { \le \,n} \right)} {\left( \matrix{ n \cr k \cr} \right)\left( {{1 \over 2}} \right)^{\,k} B_{\,n - k} } = 2^{\,n} B_{\,n} (1/2) \cr} $$

It is known that the values of Bernoulli Polynomials at $x=1/2 are given by $$ B_{\,n} (1/2) = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} $$ and therefrom the demonstration that $$ S(n) = \left( {2 - 2^{\,n} } \right)B_{\,n} $$

The demonstration of the identity for $B_{\,n} (1/2)$ descends from the multiplicative identity $$ \eqalign{ & B_{\,n} (mx) = m^{\,n - 1} \sum\limits_{k = 0}^{m - 1} {B_{\,n} (x + k/m)} \quad \Rightarrow \cr & \Rightarrow \quad B_{\,n} \left( {2 \cdot {1 \over 2}} \right) = B_{\,n} \left( 1 \right) = 2^{\,n - 1} \left( {B_{\,n} \left( {{1 \over 2}} \right) + B_{\,n} \left( {{1 \over 2} + {1 \over 2}} \right)} \right)\quad \Rightarrow \cr & \Rightarrow \quad B_{\,n} \left( {{1 \over 2}} \right) = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} \left( 1 \right) = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} \left( 0 \right) = \cr & = \left( {{1 \over {2^{\,n - 1} }} - 1} \right)B_{\,n} \quad \left| {\;0 \le n} \right. \cr} $$

and you can find a proof of the multiplicative identity in this related post, which is not .. "very complicated".