The sum $\sum_{r=0}^{20}r(20-r)\binom{20}{r}^2$

Question

Find the value of: $$\displaystyle \sum_{r=0}^{20}r(20-r)\dbinom{20}{r}^2$$

Attempt:

Using, $\dbinom{n}{r}= \dfrac n r \dbinom{n-1}{r-1}$, I get $400\sum_{r=0}^{20}\left(\dbinom {19} r \dbinom{19}{r-1}\right)$

What do I do next? I tried to use $\dbinom n r +\dbinom n {r-1}= \dbinom{n+1}{r}$ but that didn't help. Also, I know that the sum of square of binomial coefficients is $\dbinom {2n}{n}$ and the Vandermonde's identity is $\dbinom{n+m}{r}= \sum _{k=0}^r \dbinom m k \dbinom n {r-k}$.

Answer 1

Hint: Write the sum as $$\sum_{r=0}^{20}\left(r\binom{20}{r}\right)\cdot\left((20-r)\binom{20}{20-r}\right) = \sum_{r = 0}^{20}a_ra_{20-r},$$ where $a_r = r\binom{20}{r}.$ This is the coefficient of $x^{20}$ in the expansion of $$\left(\sum_{r=0}^{20}a_rx^r\right)^2.$$

Answer 2

$$\sum_{r=0}^{20} r(20-r)\binom {20}{r} ^2=20\sum_{r=0}^{20} r\binom {20}{r} ^2- \sum_{r=0}^{20} r^2\binom {20}{r} ^2=400\sum_{r=1}^{20} \binom {19}{r-1}\binom {20}{20-r} - 400\sum_{r=1}^{20} \binom {19}{r-1}\binom {19}{19-(r-1)}$$

Hence by Vandermonde's identity we get $$400\left(\sum_{r=1}^{20} \binom {19}{r-1}\binom {20}{20-r} - \sum_{r=1}^{20} \binom {19}{r-1}\binom {19}{19-(r-1)}\right) =400\left(\binom {39}{19}- \binom {38}{19}\right) $$

Now by Pascal's rule $$400\left(\binom {39}{19}- \binom {38}{19}\right) =400\binom {38}{18}$$

Alternatively you can reach answer by your way too

$$400\sum_{\color{red}{r=1}}^{\color{red}{19}} \binom {19}{r}\binom {19}{20-r}$$

Hence by using Vandermonde's identity we get $$400\sum_{r=1}^{19} \binom {19}{r}\binom {19}{20-r}=400\binom {38}{20}=400\binom {38}{18}$$