# Sum of Euler Numbers by Binomial Coefficients Vanishes/Corollary

## Theorem

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

Then:

 $\displaystyle E_{2 n}$ $=$ $\displaystyle -\sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}$ $\displaystyle$ $=$ $\displaystyle -\paren {\binom {2 n} 0 E_0 + \binom {2 n} 2 E_2 + \binom {2 n} 4 E_4 + \cdots + \binom {2 n} {2 n - 2} E_{2 n - 2} }$

where $E_n$ denotes the $n$th Euler number.

## Proof

$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$

If:

$\forall n \in \Z_{>0}: \displaystyle \sum_{k \mathop = 0}^n \binom {2 n} {2 k} E_{2 k} = 0$

Then:

 $\displaystyle \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k} + E_{2 n}$ $=$ $\displaystyle 0$ $\displaystyle E_{2 n}$ $=$ $\displaystyle - \sum_{k \mathop = 0}^{n - 1} \dbinom {2 n} {2 k} E_{2 k}$

$\blacksquare$