allocated quizz on the MO


\sum_{k=0}^{n}\frac{4^k(k+1)}{{2k\choose k}}=\frac{4^{n+1}(2n^2+5n+2)}{5{2(n+1)\choose n+1}}+\frac{1}{5}

found!

Follow the soap opera:

 

 

 

http://mathoverflow.net/questions/67784/finite-sum-of-integers-inverses

or

http://mathoverflow.net/questions/68090/finite-sums-with-binomial-and-catalan-inverses

2 Comments

Filed under math

2 responses to “allocated quizz on the MO

  1. m-k

    everyone (interested) should put in the Wolfram Alpha: sum_{k=n}^{∞}(4^k/Binomial(2k,k)) to see the tail of a funny series, he.he..

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