Bernoulli Recursions: Formula & Solutions

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SUMMARY

This discussion focuses on the recursive formulas for Bernoulli numbers, specifically highlighting the formula \sum_{k = 0}^{n-1} \binom{n}{k} B_{k} = 0. The user seeks assistance in demonstrating the equation (1 - 2^{2k}) B_{2k} = \sum_{r = 1}^{k} B_{2(k-r)} \binom{2k}{2r} (2^{2(k-r) - 1} - 1). The conversation emphasizes the importance of reliable sources, such as MathWorld, for understanding these formulas.

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yasiru89
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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,

[tex]\sum_{k = 0}^{n-1} \binom{n}{k} B_{k} = 0[/tex]
 
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One of the first places I tried, maybe it'll be better if I'm a bit more specific- I need to show that,
[tex](1 - 2^{2k}) B_{2k} = \sum_{r = 1}^{k} B_{2(k-r)} \binom{2k}{2r} (2^{2(k-r) - 1} - 1)[/tex]
(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.
 

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