It is well known that the Bernoulli numbers $B\_{k}$ for $k\in\{0,1,2,\dotsc\}$ can be generated by
\begin{equation\*}
\frac{z}{\textrm{e}^z-1}=\sum\_{k=0}^\infty B\_k\frac{z^k}{k!}=1-\frac{z}2+\sum\_{k=1}^\infty B\_{2k}\frac{z^{2k}}{(2k)!}, \quad \vert z\vert<2\pi.
\end{equation\*}
Could you please find or give a proof for the identity
\begin{equation}\label{Qi-EFinat-S-ID}\tag{BQQID}
\sum\_{j=1}^{k-1}\binom{2k}{2j}\bigl(1-2^{2j-1}-2^{2k-2j-1}\bigr)B\_{2j}B\_{2k-2j}
=\bigl(2^{2k}-1\bigr)B\_{2k}
\end{equation}
for $k\ge2$?



Ask for a proof of an identity involving the product of two Bernoulli numbers
More generally, let $B\_k(x)$, $k\ge0$, be the Bernoulli polynomials defined by the exponential generating function
\begin{equation\*}
\frac{ze^{xz}}{e^z-1}=\sum\_{k=0}^\infty B\_k(x)\frac{z^k}{k!},
\end{equation\*}
thus $B\_k=B\_k(0)$.
These satisfy the general identity (which can be proven using the definition above, simple manipulations, and comparing coefficients)
\begin{equation\*}
\sum\_{i=0}^n \binom{n}{i} B\_i(x)B\_{n-i}(y)=(1-n)B\_n(x+y)+n(x+y-1)B\_{n-1}(x+y),
\end{equation\*}
that for $x=y=0$ becomes
\begin{equation\*}
\sum\_{i=0}^n \binom{n}{i} B\_iB\_{n-i}=(1-n)B\_n-nB\_{n-1},
\end{equation\*}
and for $x=y=1/2$ becomes
\begin{equation\*}
\sum\_{i=0}^n \binom{n}{i} B\_i(1/2)B\_{n-i}(1/2)=(1-n)B\_n(1).
\end{equation\*}
Since $B\_i(1/2)=(2^{1-i}-1)B\_i$, the LHS of the last equation is
\begin{align\*}
\sum\_{i=0}^n \binom{n}{i} B\_i(1/2)B\_{n-i}(1/2)&=
\sum\_{i=0}^n \binom{n}{i} (2^{1-i}-1)B\_i\times(2^{1-n+i}-1)B\_{n-i}\\
&=\sum\_{i=0}^n \binom{n}{i} (1-2^{1-i}-2^{1-n+i}+2^{2-n})B\_iB\_{n-i}\\
&=\sum\_{i=0}^n \binom{n}{i}B\_iB\_{n-i}
+2^{2-n}\sum\_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i},
\end{align\*}
hence
\begin{equation\*} 
(1-n)B\_n(1)=(1-n)B\_n-nB\_{n-1}+2^{2-n}\sum\_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i}.
\end{equation\*}
If $n\ge4$ is even, then $B\_{n}(1)=B\_n$ and $B\_{n-1}=0$, and it follows that
\begin{align\*}
0&=\sum\_{i=0}^n \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i}\\
&=(1-2^{n})B\_n+\sum\_{i=2}^{n-2} \binom{n}{i} (1-2^{n-i-1}-2^{i-1})B\_iB\_{n-i},
\end{align\*}
which proves your identity.


