Sum with Bernoulli numbers 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!
 A: 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.
A: 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".
