I'm unsure where this should go, but because it is related to a formula for $\displaystyle \zeta(2k)$, I've put it in here. The question is regarding this recursive formula for Bernoulli numbers

$\displaystyle \displaystyle \sum^{n-1}_{k=0} {n \choose k} B_k = 0$

Essentially, we therefore have the following;

$\displaystyle \displaystyle {0 \choose k} B_0 + {1 \choose k} B_1 + {2 \choose k} B_2 + \ldots + {n-1 \choose k} B_{n-1} = 0$

Given we also have

$\displaystyle \displaystyle {n-j \choose k} = {n \choose k} \quad (\forall \,\, 1 \leq j \leq n-1)$

Then we can pair each $\displaystyle k$ for $\displaystyle 1 \leq k \leq n-1$, to obtain

$\displaystyle \displaystyle {0 \choose k} B_0 + {1 \choose k}(B_{n-1}+ B_1 ) + \ldots + {j \choose k} (B_{n-j} + B_j)= 0$

Where each $\displaystyle (n-j) + j = n$. I then wanted to consider $\displaystyle n$ even and $\displaystyle n$ odd, but I'm unsure if this is even the right way to go about. If so, how on earth do you validate that recursive formula?

Thanks a lot in advance,

HTale.