Question

Let $x+y=k$ where $x,y>0$ and
$S(k,n)={\sum }_{r=0}^n{r}^2\left({{}^{n}C}_r\right)x^r{y}^{n-r}$ then

• A. $S(1,5)=5xy+25x^2$
• B. $S(2,3)=2(3xy+9x^2)$
• C. $S(4,4)=64(xy+4x^2)$
• D. None of these

Answer 1

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

The correct options are
A $S(1,5)=5xy+25x^2$
B $S(2,3)=2(3xy+9x^2)$
C $S(4,4)=64(xy+4x^2)$

Let $x=y=1$.

We have
$r^2\left({{}^{n}C}_r\right)=\left(r\right(r-1)+r)\left({{}^{n}C}_r\right)=n(n-1)\left({{}^{n-2}C}_{r-2}\right)+n\left({{}^{n-1}C}_{r-1}\right)$
Thus, $S(1,n)=n(n-1){\sum }_{r=2}^n{{}^{n-2}C}_{r-2}{x}^r{y}^{n-r}+n{\sum }_{r=1}^n{{}^{n-1}C}_{r-1}{x}^r{y}^{n-r}=n(n-1)x^2+nx$
$=nx(nx+y)$
When $x+y=k$ we write
$S(k,n)=k^n{\sum }_{r=0}^n{r}^2{{}^{n}C}_r{\left(\frac{x}{k}\right)}^r{\left(\frac{y}{k}\right)}^{r-1}=n{k}^n\left(\frac{x}{k}\right)[n\left(\frac{x}{k}\right)+\frac{y}{k}]$
$=k^{n-2}nx(nx+y)$