Question

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

• A. $nxy$
• B. $nx(x+yn)$
• C. $nx(nx+y)$
• D. $nx$

Answer 1

The correct option is C $nx(nx+y)$

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

$\sum _{r=0}^n\left(r\right(r-1\left)\right)\cdot {}^n{C}_r{x}^r{y}^{n-r}=x^2\frac{d^2}{d{x}^2}(\sum _{r=0}^n$ ${}^n{C}_r{x}^r{y}^{n-r})=x^2\frac{d^2}{d{x}^2}(x+y{)}^n=n(n-1)x^2$
$\sum _{r=0}^{n}r\cdot {}^n{C}_r{x}^r{y}^{n-r}=x\frac{d}{dx}(\sum _{r=0}^n$ ${}^n{C}_r{x}^r{y}^{n-r})=x\frac{d}{dx}(x+y{)}^n=nx$
$\therefore \sum _{r=0}^n\left(r\right(r-1\left)\right)\cdot {}^n{C}_r{x}^r{y}^{n-r}+\sum _{r=0}^{n}r\cdot {}^n{C}_r{x}^r{y}^{n-r}$

$=n(n-1)x^2+nx$
$=nx\left[x\right(n-1)+1]$
$=nx[nx-x+(x+y\left)\right]$ ...it is given in the question that $x+y=1$
$=nx[nx+y]$
Hence, the answer is Option C