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:
$$ 2^{2n-1} - {{2n-1} \choose {n}}$$