Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient

Theorem
Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

where $B_n$ denotes the $n$th Bernoulli number.

Proof
The proof proceeds by induction.

For all $n \in \Z_{> 0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$

Basis for the Induction
$P \left({1}\right)$ is the case:

Thus $P \left({1}\right)$ is seen to hold.

This is the basis for the induction.

Induction Hypothesis
Now it needs to be shown that, if $P \left({r}\right)$ is true, where $r \ge 2$, then it logically follows that $P \left({r + 1}\right)$ is true.

So this is the induction hypothesis:
 * $\displaystyle \sum_{k \mathop = 1}^r \dbinom {2 r + 1} {2 k} 2^{2 k} B_{2 k} = 2 r$

from which it is to be shown that:
 * $\displaystyle \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k} = 2 \left({r + 1}\right)$

Induction Step
This is the induction step:

So $P \left({r}\right) \implies P \left({r + 1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k} = 2 n$