Let ${\rm E}_{2n}$ denote the even indexed Euler numbers. Prove that:
$${\rm E}_0=1, \;\; {\rm E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}{\rm E}_{2k}, \; n\geq 1$$
Solution
$${\rm E}_0=1, \;\; {\rm E}_{2n}=-\sum_{k=0}^{n-1}\binom{2n}{2k}{\rm E}_{2k}, \; n\geq 1$$
Solution
