I am trying to find an exact formula for the following:
$\sum\limits_{i=0}^{n}{\binom{2n}{n-i}\frac{2i^2+i}{n+i+1}}$
I don't think this should be too bad with a rearrangement of terms, but I keep getting stuck.
I am trying to find an exact formula for the following:
$\sum\limits_{i=0}^{n}{\binom{2n}{n-i}\frac{2i^2+i}{n+i+1}}$
I don't think this should be too bad with a rearrangement of terms, but I keep getting stuck.
Here are some steps to the solution:
Putting all these together yields that the sum $S$ to be evaluated is $ S=ns_0-(4n-1)t_1+4nt_2, $ that is, $ S=n2^{2n}-(4n-1)\tfrac12\left(2^{2n}-\textstyle{{2n\choose n}}\right)+n\left(2^{2n}-4\textstyle{{2n-1\choose n}}\right), $ and finally, $ S=\frac12\left(2^{2n}-{2n\choose n}\right)=2^{2n-1}-{2n-1\choose n}. $
Maple 16 says $ {\frac {{4}^{n} \left( \sqrt {\pi }\; \Gamma \left( n+1 \right) - \Gamma \left( n+1/2 \right) \right) }{2 \sqrt {\pi }\;\Gamma \left( n+1 \right) }} $
EDIT: Oh, looks like there's a nicer form:
$ 2^{2n-1} - {{2n-1} \choose {n}}$