Question

If ${}^n{C}_0,{}^n{C}_1,{}^n{C}_2,\cdots ,{}^n{C}_n$ denote the binomial coefficients in the expansion of $(1+x{)}^n$ and $p+q=1$ , then $\sum _{r=0}^n{r}^2{}^n{C}_r{p}^r{q}^{n-r}$ is

• A. $np$
• B. $npq$
• C. $n^2{p}^2+npq$
• D. None of these

Answer 1

The correct option is C $n^2{p}^2+npq$

We have $\sum _{r=0}^n{r}^2{}^n{C}_r{p}^r{q}^{n-r}$

$=\sum _{r=0}^n\left[r\right(r-1)+r]{}^n{C}_r{p}^r{q}^{n-r}$

$=\sum _{r=0}^{n}r(r-1){}^n{C}_r{p}^r{q}^{n-r}+\sum _{r=0}^{n}r\cdot {}^n{C}_r{p}^r{q}^{n-r}$

$=\sum _{r=0}^{n}r(r-1)\frac{n}{r}\cdot \frac{n-1}{r-1}{}^{n-2}{C}_{r-2}{p}^r{q}^{n-r}+\sum _{r=0}^{n}r\cdot \frac{n}{r}{}^{n-1}{C}_{r-1}{p}^r{q}^{n-r}$

$=n(n-1)p^2(q+p{)}^{n-2}+np(q+p{)}^{n-1}$

$=n(n-1)p^2+np[\because q+p=1]$

$=n^2{p}^2-n{p}^2+np$

$=n^2{p}^2-np(p-1)$ $[\because p-1=q]$

$=n^2{p}^2+npq[\because q+p=1]$