Q:

How can we show that $\sum_{k=1}^{n}{2n+2\choose 2k}B_{2k}=n?$

Given the sum

$$\sum_{k=1}^{n}{2n+2\choose 2k}B_{2k}=n\tag1$$
  Where $B_{2k}$ is Bernoulli number

It is quite interesting to me, the answer results in a natural number, how do you go about showing that?  I haven't got any ideas where to start. 

A:

We seek to evaluate
$$Q_n = \sum_{k=1}^n {2n+2\choose 2k} B_{2k}.$$
With this in mind we introduce the generating function
$$G(z) = \sum_{n\ge 1} Q_n \frac{z^{2n+2}}{(2n+2)!}.$$
We then obtain for $G(z)$ that
$$G(z) = \sum_{n\ge 1} \frac{z^{2n+2}}{(2n+2)!} 
\sum_{k=1}^n {2n+2\choose 2k} B_{2k}
\\ = \sum_{n\ge 1} \frac{z^{2n+2}}{(2n+2)!} 
\sum_{k=1}^n {2n+2\choose 2k} 
(2k)! [ w^{2k}] \frac{w}{\exp(w)-1}
\\ = \sum_{n\ge 1} z^{2n+2}
\sum_{k=1}^n \frac{1}{(2n+2-2k)!} 
[w^{2k}] \frac{w}{\exp(w)-1}
\\ = \sum_{k\ge 1} [w^{2k}] \frac{w}{\exp(w)-1}
\sum_{n\ge k} z^{2n+2} \frac{1}{(2n+2-2k)!} 
\\ = \sum_{k\ge 1} z^{2k} [w^{2k}] \frac{w}{\exp(w)-1}
\sum_{n\ge 0} z^{2n+2} \frac{1}{(2n+2)!} 
\\ = \left(-1 + \frac{1}{2}(\exp(z)+\exp(-z))\right)
\sum_{k\ge 1} z^{2k} [w^{2k}] \frac{w}{\exp(w)-1}
\\ = \left(-1 + \frac{1}{2}(\exp(z)+\exp(-z))\right)
\left(-1 + \frac{1}{2} z + \frac{z}{\exp(z)-1}\right).$$
As we examine this formula we  see that it could have been obtained by
inspection.   We put $\exp(z) = v$ while we simplify and get
$$\frac{1}{v-1}
\left(-1 + \frac{1}{2}v+\frac{1}{2}\frac{1}{v}\right)
\left(1-v + \frac{1}{2}zv - \frac{1}{2}z+z\right)
\\ = \frac{1}{2}\frac{1}{v}\frac{1}{v-1}
(-2v+v^2+1)\left(1-v+\frac{1}{2}zv+\frac{1}{2}z\right)
\\ = \frac{1}{2}\frac{1}{v} (v-1)
\left(1-v+\frac{1}{2}zv+\frac{1}{2}z\right)
\\ = \frac{1}{2}\frac{1}{v}
\left(-1+v-\frac{1}{2}zv-\frac{1}{2}z
+ v-v^2+\frac{1}{2}zv^2+\frac{1}{2}zv\right)
\\ = \frac{1}{2}\frac{1}{v}
\left(-1+2v-\frac{1}{2}z
-v^2+\frac{1}{2}zv^2\right)
\\ = \frac{1}{2}
\left(-\frac{1}{v}+2-\frac{1}{2}z\frac{1}{v}
-v+\frac{1}{2}zv\right)$$
Extracting coefficients from this now yields
$$(2n+2)! [ z^{2n+2}] \frac{1}{2}
\left(-\frac{1}{v}+2-\frac{1}{2}z\frac{1}{v}
-v+\frac{1}{2}zv\right)
\\ = \frac{1}{2}
\left(-(-1)^{2n+2} - \frac{1}{2} (2n+2) (-1)^{2n+1}
- 1 + \frac{1}{2} (2n+2) \right) 
\\ = \frac{1}{2}
\left(-2 + (n+1) + (n+1)\right).$$
We thus have
$$\bbox[5px,border:2px solid #00A000]{Q_n = n.}$$
Observe  that we may  verify that  the coeffcients  of odd  powers are
zero.  We obtain
$$(2n+1)! [ z^{2n+1}] \frac{1}{2}
\left(-\frac{1}{v}+2-\frac{1}{2}z\frac{1}{v}
-v+\frac{1}{2}zv\right)
\\ = \frac{1}{2}
\left(-(-1)^{2n+1} - \frac{1}{2} (2n+1) (-1)^{2n}
- 1 + \frac{1}{2} (2n+1)\right)
\\ = \frac{1}{2}
\left(- \frac{1}{2} (2n+1)
+ \frac{1}{2} (2n+1)\right)
= 0.$$
The same kind of arithmetic works for $[z^0].$
Remark.  The  above used the technique  of annihilated coefficient
extractors (ACE), otherwise known as the substitution rule for formal
power  series.   There  are several  more  examples at  this MSE  link
I  and at  this MSE
link  II  and  also
here             at             this             MSE             link
III.