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In a recent failed-post about some partial sums with respect to the Central Binomial and Catalan number the formulas $$\sum_{k=0}^n\frac{4^k}{B_k}=\frac{4^{n+1}(2n+1)}{3 B_{n+1}}+\frac{1}{3}$$ $$\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{4^{n+1}(2n^2+5n+2)}{5 B_{n+1}}+\frac{1}{5}$$ were mention, here in MO, and I forgot to ask, so let me do it now:

Is This just one instance of some broader well known pattern?

Here, consider $B_m={2m \choose m}$ and $C_m=\frac{B_m}{m+1}$ for those set of numbers.

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1 Answer 1

The lhs and rhs in the first formula differ on $-\frac{4 \sqrt{\pi } \Gamma (n+2)}{3 \Gamma \left(n+\frac{1}{2}\right)}$. For example, then $n=0$ we have $lhs-rhs=-4/3$. But it can be mended.

A natural way to approach this sort of equalities are generating functions. Let $$ f(x)=\sum_{n=0}^\infty \frac{(4x)^n}{B_n}= \frac{1}{1-x}+\frac{\sqrt{x} \sin ^{-1}\left(\sqrt{x}\right)}{(1-x)^{3/2}}, $$ $$ A_n=\sum_{k=0}^n \frac{4^k}{B_k}. $$ Then
$$ \sum_{n=0}^\infty A_n x^n=\frac{f(x)}{1-x}, $$ $$ f'(x)=\sum_{n=0}^\infty \frac{4^{n+1}(n+1)x^n}{B_{n+1}},$$ $$\sum_{n=0}^\infty \frac{4 \sqrt{\pi } \Gamma (n+2)}{3 \Gamma \left(n+\frac{1}{2}\right)}x^n= ((2 x+(6 \sqrt{x} sin^{-1}(\sqrt{x}))/\sqrt{1-x}+4)/(3 (x-1)^2)).$$ And we indeed have $$ \frac{f(x)}{1-x}- 2f'(x)+\left(\frac{f(x)}{x}-\frac{1}{x}\right)-\frac1{3(1-x)}=-((2 x+(6 \sqrt{x} sin^{-1}(\sqrt{x}))/\sqrt{1-x}+4)/(3 (x-1)^2)). $$

I think it is possible to obtain formulas for $\sum_{k=0}^n \frac{4^k k^m}{B_k}$, $m\in\mathbb N$ analogously.

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you are right, i make a mistake because i should write the next formulas: $\sum_{k=0}^n\frac{4^k}{B_k}=\frac{4^{n+1}(2n+1)}{3B_{n+1}}+\frac{1}{3}$ and $\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{4^{n+1}(2n^2+5n+2)}{5B_{n+1}}+\frac{1}{5‌​}$ –  janmarqz Jun 18 '11 at 1:04

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