I wanted to find the value of the following expression

$$\sum_{r=0}^{20} (-1)^r. \binom {30}{r}. \binom {30}{r+10}$$

The Vandermonde's identity could have been used in this problem but the $(-1)^r$ makes it of no use. Using the binomial expansion and some brute force I got the answer as the coefficient of $x^{20}$ in the expansion $(1-x^2)^{30}$. But I am not able to provide the proof in writing because it was just by brute force and some observation. Can someone please provide a hint.

  • $\begingroup$ Mathematica gives the answer immediately: $30045015$. What do you mean "closed form" in this context? $\endgroup$ – David G. Stork Jan 24 '18 at 18:24
  • $\begingroup$ @David G. Stork I want a written mathematical proof for my obtained answer not just the expansion of $\binom {30}{10}$. Also typo fixed. $\endgroup$ – Rohan Shinde Jan 24 '18 at 18:26
  • $\begingroup$ What about writing two terms, one for all the $(-1)^{+1}$ cases, and one for all the $(-1)^{-1}$ cases, use identities, and then subtract the results? Hint: $2095922267925255 - 2095922237880240 = 30045015$ $\endgroup$ – David G. Stork Jan 24 '18 at 18:29
  • $\begingroup$ Hint: consider the coefficient of $x^{20}$ in $(1-x)^{30}(1+x)^{30}$. $\endgroup$ – Greg Martin Jan 24 '18 at 18:30
  • $\begingroup$ @Greg Martin how did you arrive at that result. Because somewhere in my roughwork I saw that coming but now I don't remember how I got that one. $\endgroup$ – Rohan Shinde Jan 24 '18 at 18:35

It is convenient to use the coefficient of operator $[z^n]$ to denote the coefficient of $z^n$ in a series. This way we can write for instance \begin{align*} [z^k](1+z)^n=\binom{n}{k} \end{align*}

We obtain \begin{align*} \color{blue}{\sum_{r=0}^{20}}&\color{blue}{(-1)^r \binom {30}{r} \binom {30}{r+10}}\\ &=\sum_{r=0}^{20}(-1)^r \binom {30}{30-r} \binom {30}{r+10}\tag{1}\\ &=\sum_{r\geq 0}[z^{30-r}](1-z)^{30}[u^{r+10}](1+u)^{30}\tag{2}\\ &=[z^{30}](1-z)^{30}\sum_{r\geq 0}z^r[u^r]u^{-10}(1+u)^{30}\tag{3}\\ &=[z^{30}](1-z)^{30}z^{-10}(1+z)^{30}\tag{4}\\ &=[z^{40}](1-z^2)^{30}\tag{5}\\ &=[z^{40}]\sum_{k=0}^{30}\binom{30}{k}(-1)^kz^{2k}\\ &\color{blue}{=\binom{30}{20}=30\,045\,015}\tag{6} \end{align*}


  • In (1) we use the binomial identity $\binom{p}{q}=\binom{p}{p-q}$.

  • In (2) we apply the coefficient of operator twice. We also set the upper limit to $\infty$ without changing anything since we are adding zeros only.

  • In (3) we use the linearity of the coefficient of operator and we apply the rule $[z^{p-q}]A(z)=[z^p]z^qA(z)$.

  • In (4) we apply the substitution rule of the coefficient of operator with $u=z$
    \begin{align*} A(z)=\sum_{r\geq0} a_r z^r=\sum_{r\geq 0} z^r [u^r]A(u) \end{align*}

  • In (5) we do some simplifications and apply the same rule as we did in (3).

  • In (6) we select the coefficient of $z^{40}$.


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