Question

If $x+y=1$, then $\sum _{r=0}^n{r}^{2\cdot n}{C}_r{x}^r{y}^{n-r}$ is equal to

• A. nxy
• B. nx (x +yn)
• C. nx (n x +y)
• D. None of these

Answer 1

Answer: (C)

$\sum _{r=0}^n{r}^2{}^n{C}_r{x}^r{y}^{n-r}$
$=\sum _{r=0}^n[r(r-1)+r]{}^n{C}_r{x}^r{y}^{n-r}$
$=\sum _{r=0}^{n}r(r-1){}^n{C}_r{x}^r{y}^{n-r}$
$+\sum _{r=0}^n{r}^n{C}_r{x}^r{y}^{n-r}$
$=\sum _{r=2}^{n-2}r(r-1)\frac{n}{r}\cdot \frac{n-1}{r-1}{}^{n-2}{C}_{r-2}{x}^2\cdot x^{r-2}{y}^{n-r}$
$+\sum _{r=1}^{n}r\cdot \frac{n}{r}{}^{n-1}{C}_{r-1}x\cdot x^{r-1}{y}^{n-r}$
$=n(n-1)x^2\sum _{r=2}^{n-2}{}^{n-2}{C}_{r-2}{x}^{r-2}{y}^{(n-2)-(r-2)}$
$+nx\sum _{r=1}^n{}^{n-1}{C}_{r-1}{x}^{r-1}{y}^{(n-1)-(r-1)}$
$=n(n-1)x^2(x+y{)}^{n-2}+nx(x+y{)}^{n-1}$
$=n(n-1)x^2+nx(\because x+y=1)$
$=nx(nx-x+1)$
$=nx(nx+y)(\because x+y=1)$