# Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient

## Theorem

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

Then:

 $\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}$ $=$ $\ds \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots$ $\ds$ $=$ $\ds 2 n$

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

## Proof 1

Let $B_k$ denote the kth Bernoulli number

Let $B_k \left(x\right)$ denote the kth Bernoulli polynomial

 $\ds B_{2 n + 1} \left({\frac 1 2}\right)$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds 2^{2 n + 1} \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{-k}$ $=$ $\ds 0$ $\ds \leadsto \ \$ $\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k$ $=$ $\ds 0$
 $\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k$ $=$ $\ds \left({2 n + 1}\right) 2 B_1 + \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}$ $\ds$ $=$ $\ds -\left({2 n + 1}\right) + \sum_{k \mathop = 0}^n \binom {2 n + 1 } {2 k} B_{2 k} 2^{2 k}$ $B_1 = -\frac 1 2$ $\ds \leadsto \ \$ $\ds \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}$ $=$ $\ds 2 n + 1$ $\ds \leadsto \ \$ $\ds \sum_{k \mathop = 1}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}$ $=$ $\ds 2 n + 1 - B_0$ $\ds$ $=$ $\ds 2 n$ $B_0 = 1$

$\blacksquare$

## Proof 2

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:

 $\ds \binom {2 \times 1 + 1} 2 2^2 B_2$ $=$ $\ds \frac {2 \times 3} 2 \times 2^2 \times \frac 1 6$ Binomial Coefficient with Two, Definition of Bernoulli Numbers $\ds$ $=$ $\ds 2$ after simplification $\ds$ $=$ $\ds 2 \times 1$

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:

 $\ds \sum_{k \mathop = 1}^{r + 1} \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k}$ $=$ $\ds \sum_{k \mathop = 1}^r \dbinom {2 \left({r + 1}\right) + 1} {2 k} 2^{2 k} B_{2 k} + \dbinom {2 \left({r + 1}\right) + 1} {2 \left({r + 1}\right)} 2^{2 \left({r + 1}\right)} B_{2 \left({r + 1}\right)}$ $\ds$ $=$ $\ds \sum_{k \mathop = 1}^r \dbinom {2 r + 3} {2 k} 2^{2 k} B_{2 k} + \left({2 r + 3}\right) 2^{2 r + 2} B_{2 r + 2}$ Binomial Coefficient with Self minus One and simplification

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

Therefore:

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