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I am trying to prove the following for a very, very long time: $ 2k=\sum_{j=1}^k \binom{2k+1}{2j}2^{2j}B_{2j} $ Here, $B_{2j}$ are Bernoulli numbers.

I would be extremely happy if somebody could help me with this!

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    The following [MSE link](http://math.stackexchange.com/questions/822466/sum-involving-bernoulli-numbers-sum-r-1n-binom2n2r-1-fracb-2rr) points to a computation that can easily be adapted to prove the result being asked for here.2014-08-06

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Let $B_n(X) := \sum_{i=0}^n \binom{n}{i}B_i X^{n-i}$ denote the $n$-th Bernoulli polynomial. It satisfies the symmetry relation $B_n(1-X) = (-1)^n B_n(X)$ which implies $B_n(1/2) = 0$ for odd numbers $n$. Hence $0 = 2^{2k+1} B_{2k+1}(1/2) = \sum_{i=0}^{2k+1} \binom{2k+1}{i} B_i 2^i = 1 + (2k+1)2B_1 + \sum_{i=2}^{2k+1} \binom{2k+1}{i} B_i 2^i.$ All summands with odd $i$ vanish since the odd-numbered Bernoulli numbers are zero, the only exception being $B_1 = -1/2$. It follows that $0 = -2k + \sum_{j=1}^k \binom{2k+1}{2j}B_{2j}2^{2j}$ which proves your proposition.