Question

For $n\in N$. let
$S\left(k\right)={\sum }_{r=0}^n{r}^k{{(C}_r)}^2$ then

• A. $S\left(0\right)={{}^{2n}C}_n$
• B. $S\left(1\right)=\frac{1}{2}n\left({{}^{2n}C}_n\right)$
• C. $S\left(2\right)=n^2\left[\frac{1}{2}\right({{}^{2n}C}_n)-{{}^{2n-2}C}_{n-2}]$
• D. $S\left(0\right)+S\left(1\right)+S\left(2\right)=2^{2n-3}\left({{}^{2n}C}_n\right)$

Answer 1

Answer: (A), (B), (C)

$S\left(0\right)={{}^{2n}C}_n$
$S\left(2\right)=n^2\left[\frac{1}{2}\right({{}^{2n}C}_n)-{{}^{2n-2}C}_{n-2}]$
$S\left(1\right)=\frac{1}{2}n\left({{}^{2n}C}_n\right)$

$S\left(0\right)={\sum }_{r=0}^n{\left({{}^{n}C}_r\right)}^2={\sum }_{r=0}^n{\left({{}^{n}C}_r\right)}^{}({\sum }_{r=0}^n{\left({{}^{n}C}_{n-r}\right)}^{}$
$=$ number of ways of selecting $n$ persons out of $n$ men and $n$ women
$={{}^{2n}C}_n$
$S\left(1\right)={\sum }_{r=0}^n{r\left({{}^{n}C}_r\right)}^2={\sum }_{r=0}^n{r\left({{}^{n}C}_{n-r}\right)}^2={\sum }_{r=0}^n{(n-r)\left({{}^{n}C}_r\right)}^2=n\left({{}^{2n}C}_n\right)-S\left(1\right)$
$\Rightarrow S\left(1\right)=\frac{n}{2}\left({{}^{2n}C}_n\right)$
$S\left(2\right)={\sum }_{r=0}^n{r\left({{}^{n}C}_r\right)}^2={\sum }_{r=0}^n{r[n-(n-r\left)\right]\left({{}^{n}C}_r\right)}^2$
$=n{\sum }_{r=0}^n{r\left({{}^{n}C}_r\right)}^2-{\sum }_{r=0}^n{r\left({{}^{n}C}_r\right)}^2\left(\right(n-r\left){{}^{n}C}_{n-r}\right)$
But
${\sum }_{r=0}^n{\left(r{{}^{n}C}_r\right)}^{}\left(\right(n-r\left){{}^{n}C}_{n-r}\right)$
$=$ the number of ways of selecting $n$ persons from $n$ men and $n$ women and appointing a men's leader and a women's leader
$=n^2\left({{}^{2n-2}C}_{n-2}\right)$
Thus,
$S\left(2\right)=\frac{n^2}{2}\left({{}^{2n}C}_n\right)-n^2\left({{}^{2n-2}C}_{n-2}\right)$