Question

If $x+y=1,x\ne \{0,1\}$, then for which value of $x$, $\sum _{r=0}^{10}{r}^2\left({}^{10}{C}_r\right)x^r{y}^{10-r}=0$

• A. $9$
• B. $-9$
• C. $\frac{1}{9}$
• D. $-\frac{1}{9}$

Answer 1

The correct option is D $-\frac{1}{9}$

$\sum _{r=0}^n{r}^2\left({}^n{C}_r\right)x^r{y}^{n-r}=\sum _{r=0}^n\left(r\right(r-1)+r)\left({}^n{C}_r\right)x^r{y}^{n-r}=\sum _{r=0}^{n}r(r-1)\left({}^n{C}_r\right)x^r{y}^{n-r}+\sum _{r=0}^{n}r\left({}^n{C}_r\right)x^r{y}^{n-r}=\sum _{r=2}^{n}n(n-1){}^{n-2}{C}_{r-2}{x}^2{x}^{r-2}{y}^{n-r}+\sum _{r=1}^{n}n{}^{n-1}{C}_{r-1}x{x}^{r-1}{y}^{n-r}(\because r{}^n{C}_r=n{}^{n-1}{C}_{r-1}\&r(r-1){}^n{C}_r=n(n-1\left){}^{n-2}{C}_{r-2}\right)=n(n-1)x^2\sum _{r=2}^n{}^{n-2}{C}_{r-2}{x}^{r-2}{y}^{n-r}+nx\sum _{r=1}^n{}^{n-1}{C}_{r-1}{x}^{r-1}{y}^{n-r}=n(n-1)x^2(x+y{)}^{n-2}+nx(x+y{)}^{n-1}=n(n-1)x^2+nx\therefore \sum _{r=0}^{10}{r}^2\left({}^n{C}_r\right)x^r{y}^{n-r}=010\times 9x^2+10x=010x(9x+1)=0\Rightarrow x=-\frac{1}{9}$