# Sum of Binomial coefficients identity

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.

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WA doesn't seem to give a nice formula... –  user17762 Nov 29 '12 at 7:16
But that is without the other part of the sum and infinite..... –  Joe Nov 29 '12 at 7:18
Ah, never mind, the link wasn't working properly at first. I have been told there is a nice exact form though –  Joe Nov 29 '12 at 7:23

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:

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

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+1 -- interesting -- surprisingly small sequence number at OEIS. –  joriki Nov 29 '12 at 8:06

Here are some steps to the solution:

• To get rid of the denominators, use ${2n\choose n-i}\frac1{n+i+1}={2n+1\choose n-i}\frac1{2n+1}$.
• To simplify the steps to come, use the change of variable $k=n-i$, thus ${2n+1\choose n-i}(2i^2+i)={2n+1\choose k}((2n+1)n-(4n-1)k+2k(k-1))$ and $0\leqslant k\leqslant n$.
• To compute $s_0=\sum\limits_{k=0}^n{2n+1\choose k}$, note that $2s_0=\sum\limits_{k=0}^{2n+1}{2n+1\choose k}=2^{2n+1}$.
• To compute $s_1=\sum\limits_{k=0}^n{2n+1\choose k}k$, note that ${2n+1\choose k}k={2n\choose k-1}(2n+1)$ hence $s_1=(2n+1)t_1$ with $t_1=\sum\limits_{k=0}^{n-1}{2n\choose k}$ and $2t_1+{2n\choose n}=\sum\limits_{k=0}^{2n}{2n\choose k}=2^{2n}$.
• To compute $s_2=\sum\limits_{k=0}^n{2n+1\choose k}k(k-1)$, note that ${2n+1\choose k}k(k-1)={2n-1\choose k-2}(2n+1)(2n)$ hence $s_2=(2n+1)2nt_2$ with $t_2=\sum\limits_{k=0}^{n-2}{2n-1\choose k}$ and $2t_2+2{2n-1\choose n}=\sum\limits_{k=0}^{2n-1}{2n-1\choose k}=2^{2n-1}$.

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, $$S=\frac12\left(2^{2n}-{2n\choose n}\right)=2^{2n-1}-{2n-1\choose n}.$$

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