Originally Posted by

**chisigma** If you remember the explicit formula for the Bernuoulli number of index m...

$\displaystyle B_{m}= \sum_{n=0}^{m} \frac{1}{n+1}\cdot \sum_{k=0}^{n} (-1)^{k}\cdot \binom{n}{k}\cdot k^{m}$

... you can obtain...

$\displaystyle - \frac{1}{6}\cdot \sum_{k=0}^{5}(-1)^{k}\cdot \binom{5}{k}\cdot k^{5} =B_{5} - \sum_{n=0}^{4}\frac{1}{n+1}\cdot \sum_{k=0}^{n}(-1)^{k}\cdot \binom{n}{k}\cdot k^{5} $

... and you are able to verify the first identity...

For the second identity i need some more time...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$