Bernoulli Recursions: Formula & Solutions

[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

 

[Gib Z]

http://mathworld.wolfram.com/BernoulliNumber.html

Thats got quite a few, and due to trust of the source, definitely almost every single elementary one.

 

[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.