# Bernoulli Recursions

1. Jan 10, 2008

### yasiru89

I don't know if this is appropriate or not, here or anywhere. However, I propose this thread be used to post recursive formula for the Bernoulli numbers. It saves a great deal of frustration.
The first is simply,

$$\sum_{k = 0}^{n-1} \binom{n}{k} B_{k} = 0$$

2. Jan 11, 2008

### Gib Z

3. Jan 11, 2008

### yasiru89

One of the first places I tried, maybe it'll be better if I'm a bit more specific- I need to show that,
$$(1 - 2^{2k}) B_{2k} = \sum_{r = 1}^{k} B_{2(k-r)} \binom{2k}{2r} (2^{2(k-r) - 1} - 1)$$
(should be right...)
Tried a few approaches, didn't work, I might not be trying hard enough(bit of a hectic period), I'd sure appreciate some help! A nod in the right direction even.