Sum of Bernoulli Numbers by Power of Two and Binomial Coefficient/Proof 1

Theorem

Let $n \in \Z_{>0}$ be a (strictly) positive integer.

Then:

\(\ds \sum_{k \mathop = 1}^n \dbinom {2 n + 1} {2 k} 2^{2 k} B_{2 k}\)

\(=\)

\(\ds \binom {2 n + 1} 2 2^2 B_2 + \binom {2 n + 1} 4 2^4 B_4 + \binom {2 n + 1} 6 2^6 B_6 + \cdots\)

\(\ds \)

\(=\)

\(\ds 2 n\)

where $B_n$ denotes the $n$th Bernoulli number.

Proof

Let $B_k$ denote the kth Bernoulli number

Let $B_k \left(x\right)$ denote the kth Bernoulli polynomial

By Value of Odd Bernoulli Polynomial at One Half:

\(\ds B_{2 n + 1} \left({\frac 1 2}\right)\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds 2^{2 n + 1} \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{2 n + 1 - k}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k \left({\frac 1 2}\right)^{-k}\)

\(=\)

\(\ds 0\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k\)

\(=\)

\(\ds 0\)

By Odd Bernoulli Numbers Vanish:

\(\ds \sum_{k \mathop = 0}^{2 n + 1} \binom {2 n + 1} k B_k 2^k\)

\(=\)

\(\ds \left({2 n + 1}\right) 2 B_1 + \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}\)

\(\ds \)

\(=\)

\(\ds -\left({2 n + 1}\right) + \sum_{k \mathop = 0}^n \binom {2 n + 1 } {2 k} B_{2 k} 2^{2 k}\)

$B_1 = -\frac 1 2$

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 0}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}\)

\(=\)

\(\ds 2 n + 1\)

\(\ds \leadsto \ \ \)

\(\ds \sum_{k \mathop = 1}^n \binom {2 n + 1} {2 k} B_{2 k} 2^{2 k}\)

\(=\)

\(\ds 2 n + 1 - B_0\)

\(\ds \)

\(=\)

\(\ds 2 n\)

$B_0 = 1$

$\blacksquare$