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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?

Post Closed as "not a real question" by Steve Huntsman, Andres Caicedo, Yemon Choi, Aaron Meyerowitz, Gerry Myerson

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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