For even positive integer n find the following sum : $\binom{n}{0}^{2}-2\binom{n}{1}^{2}+3\binom{n}{2}^{2}+...+(-1)^{n}(n+1)\binom{n}{n}^{2}$

Here's what i did: Closed form of the series is $\sum_{r=0}^{n}(-1)^{r}(r+1)\binom{n}{r}^{2}$ which is equal to $\sum_{r=0}^{n}(-1)^{r}r\binom{n}{r}^{2}+\sum_{r=0}^{n}(-1)^{r}\binom{n}{r}^{2}$. For even n, the second sum is the coefficient of $x^{n}$ in $(1-x^{2})^{n}$ which is $(-1)^{n/2}\binom{n}{n/2}$. The first one (and there's my doubt) can be written as $\sum_{r=0}^{n}(-1)^{r}\binom{n-1}{r-1}^{2}$ using $\binom{n}{r}$=$\frac{n}{r}\binom{n-1}{r-1}$and then the sum is the coefficient of $x^{n-1}$ in $(1-x^{2})^{n-1}$ which is 0 for even n as $\binom{n}{r}$ is defined for non negative integers. My answer doesn't match with the book. Where am I wrong?

  • $\begingroup$ Could you post the answer given in the book ? $\endgroup$ – Claude Leibovici Oct 18 '17 at 6:09
  • $\begingroup$ The actual problem asks for the product of that sum with $\frac{2}{\binom{n}{n/2}}$ and the answer is $(-1)^{n/2}(n+2)$ $\endgroup$ – Epsilon zero Oct 18 '17 at 6:41
  • $\begingroup$ I am very bad in this area (just as in many other); so, I just computed a few values. The result given in the book is correct only if $n$ is even. $\endgroup$ – Claude Leibovici Oct 18 '17 at 7:21

For $n=0,1$ the answer is trivial, so with $n>1$ $$(1+e^{ix})^n=\sum_{k=0}^n{n\choose k}e^{ikx}\tag{1}$$ and it's derivative is $$\dfrac{d}{dx}(1+e^{ix})^ne^{ix}=\sum_{k=0}^n{n\choose k}(i)(k+1)e^{i(k+1)x}$$ or $$(1+e^{ix})^{n-1}(ne^{ix}+e^{ix}+1)=\sum_{k=0}^n{n\choose k}(k+1)e^{ikx}\tag{2}$$ multiply $(1)$ and $(2)$ and then integrate from $0$ to $2\pi$ : $$ \int_0^{2\pi}(1+e^{ix})^{n-1}(ne^{ix}+e^{ix}+1)(1-e^{-ix})^{n}\,dx = \int_0^{2\pi}\sum_{k=0}^n{n\choose k}(k+1)e^{ikx}\sum_{\ell=0}^n{n\choose \ell}(-1)^\ell e^{-i\ell x}\,dx $$ every factor $\int_0^{2\pi}e^{ikx}e^{-i\ell x}\,dx=0$ for $k\neq\ell$ and $\int_0^{2\pi}e^{ikx}e^{-i\ell x}\,dx=2\pi$ for $k=\ell$ then $$I=\sum_{k=0}^n{n\choose k}^2(-1)^k(k+1)2\pi$$ where \begin{align} I &= \int_0^{2\pi}(1+e^{ix})^{n-1}(ne^{ix}+e^{ix}+1)(1-e^{-ix})^{n}\,dx \\ &= (2i)^{n-1}n\int_0^{2\pi}\sin^{n-1}x\cos x\,dx + (2i)^{n-1}(n+2)i\int_0^{2\pi}\sin^nx\,dx - (2i)^{n-1}n\int_0^{2\pi}\sin^{n-1}x\,dx \end{align} the first integral is zero and for integer $n$ one of next integrals is zero as well. For our purpose $n=2k$ we see $\displaystyle\int_0^{2\pi}\sin^{2k-1}x\,dx=0$ and $$\int_0^{2\pi}\sin^{2k}x\,dx=\int_0^{2\pi}\left(\dfrac{e^{ix}-e^{-ix}}{2i}\right)^{2k}\,dx=\int_0^{2\pi}(-1)^k{2k \choose k}\left(\dfrac{1}{2i}\right)^{2k}\,dx=\dfrac{(-1)^k}{(2i)^{2k}}{2k \choose k}2\pi$$ which concludes $$\sum_{k=0}^n{n\choose k}^2(-1)^k(k+1)=\color{blue}{(-1)^k(k+1){2k \choose k}}$$

  • $\begingroup$ Very nice way, for sure ! $\to +1$ $\endgroup$ – Claude Leibovici Oct 18 '17 at 17:56
  • $\begingroup$ Thank you. Unfortunately it has an error and doesn't match with primary series and I couldn't find the problem! $\endgroup$ – Nosrati Oct 18 '17 at 18:00
  • $\begingroup$ After your comment, I added some results obtained using a CAS, hoping that they could (I hope) give you some ideas. Cheers. $\endgroup$ – Claude Leibovici Oct 19 '17 at 2:42

This is definitely not an answer.

After MyGlasses's comment to his/her own answer, I cheated (using a CAS) and got as a result $$ \sum_{r=0}^{n}(-1)^{r}(r+1)\binom{n}{r}^{2}=\sqrt{\pi } \,\frac{2^n}{\Gamma \left(\frac{1-n}{2}\right) \Gamma \left(\frac{n+2}{2}\right)}-n^2 \, _2F_1(1-n,1-n;2;-1)$$ Now, computing for different even values of $n$ $$a_n=\frac{2 \sum_{r=0}^{n}(-1)^{r}(r+1)\binom{n}{r}^{2}}{\binom{n}{\frac{n}{2}} }$$ we effectively obtain the sequence $$\{+2,-4,+6,-8,+10,-12,+14,-16,+18,-20,+22,\cdots\}$$ which corresponds to the answer given in the book $(-1)^{n/2}(n+2)$.

  • $\begingroup$ I appreciate for you attention (+1). I saw your answer and forced me to review my post, indeed $_2F_1(1-n,1-n;2;-1)=0$ for integer $n$ and remains the first term in our purpose. Thanks a lot! $\endgroup$ – Nosrati Oct 19 '17 at 6:48
  • $\begingroup$ @MyGlasses. You are very welcome ! $\endgroup$ – Claude Leibovici Oct 19 '17 at 6:51
  • $\begingroup$ Well this is a 9th grade problem so I guess this method is too complicated. Elementary elegant solutions any? $\endgroup$ – Epsilon zero Oct 19 '17 at 12:07
  • $\begingroup$ @AnuranChowdhury. I would really like to learn something simpler ! Please, let me know when you get the answer (may be, you could update your post with the proposed steps toward the solution). This would be of interest to many people here. Cheers. $\endgroup$ – Claude Leibovici Oct 19 '17 at 15:07

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