## Finite sum of integers inverses [closed]

Recently, we learned from Renzo Sprugnoli that $\sum_{k=0}^n \frac{4^k}{B_k} =\frac{2n+1}{3}\frac{4^{n+1}}{B_{n+1}}+\frac{1}{3}$, where $B_n$ are the famous central binomial coefficients, $B_n={2n \choose n}$.

Nowadays it is posible to find; using W|A, The Wolframalpha Calculator, that:

$$\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{2n^2+5n+2}{5}\frac{4^{n+1}}{B_{n+1}}+\frac{1}{5}.$$

My "begs" are for someone to help me to pursuit some interesant generalizations or applications about the involved Catalan numbers.

Ref: R. Sprugnoli, "Sum of reciprocals of the central binomial coefficients", INTEGERS: Electronic journal of combinatorial number theory 6 (2006), #A27.

Update:

Is the formula $\sum_{k=0}^n\frac{4^k(k+1)}{B_k}=\frac{2n^2+5n+2}{5}\frac{4^{n+1}}{B_{n+1}}+\frac{1}{5}$ well known?

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What expression is supposed to be in the summation symbols? – Qiaochu Yuan Jun 14 2011 at 17:58
Could you please try to formulate a concrete question. At the moment for me this reads like a request for collaboration or general guidance; posting such is not the purpose of this site. – quid Jun 14 2011 at 18:14
Have you looked it from the A=B (www.math.upenn.edu/~wilf/AeqB.html) point of view? (For the record, I agree with the previous commenter, there is no well posed <i>question</i> here.) – Vladimir Dotsenko Jun 14 2011 at 20:11
joan, I was responding to "My "begs" are for someone to help me to pursuit some interesant generalizations or applications about the involved Catalan numbers." This struck me as an inappropriate request for this site. I fail to see how you know that W|A can solve my theorems, but I suppose if Wolfram is to be believed then he holds the key to understanding everything... – Yemon Choi Jun 15 2011 at 3:26
@joan, comments on closed questions are unlikely to attract much attention. Better to pose a new question (perhaps after checking the faq for suggestions on the posing of questions on this site). – Gerry Myerson Jun 15 2011 at 23:33
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