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The Bernoulli numbers are the rational numbers $B_n$ defined as the coefficients in the expansion $\frac{x}{e^x-1} = \sum_{n \geq 0} B_n \frac{x^n}{n!}$. They vanish when $n$ is odd and greater than $2$. They appear in the values at integers of the Riemann $\zeta$ function. These classical numbers play an important role in number theory and in several other places in mathematics.
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Prove $ \sum_{i=0}^{2a+1} {2a+1 \choose i} B_{2a+1-i} [ (n+1)^i+(-n)^i ] =0 $ for Bernoulli ...
Dividing your expression by $(2a+1)!$ and using the definition of binomial coefficients, we see that you would like to prove that
$$
\sum_{i=0}^{2a+1} \frac{1}{(2a+1-i)!i!}B_{2a+1-i}((n+1)^i+(-n)^i)=0 …