Question

If p + q = 1 then show that ${\sum }_{r=0}^n{r}^2{c}_r{p}^r\cdot q^{n-r}=npq+n^2{p}^2$

Answer 1

${\sum }_{r=0}^n{r}^2{C}_r{p}^r{q}^{n-r}=0+1^2{C}_{1}p{q}^{n-1}+2^2{C}_2{p}^2{q}^{n-2}+....+n^2{C}_n{P}^n$....(1)
Because of factor $2^2{C}_2$ we follow the procedure of 3(a)
We have done above in Q3(a) R(2) ,
$C_1+2^2{C}_{2}x+3^2{C}_3{x}^2+....+n^2{C}_n{x}^{n-1}$
$=n(1+x{)}^{n-2}[1+x+x(n-1)]$
$=n(1+x{)}^{n-2}(1+nx)$ ....(2)
Comparing (1) and (2) and keeping in view
that $p+q$ , we put $x=p/q$ in both sides of (1)
$C_1+2^2{C}_2\frac{p}{q}+3^2{C}_3\frac{p^2}{q^2}+....+n^2{C}_n\frac{p^{n-1}}{q^{n-1}}$
$=\frac{(p+q{)}^{n-2}}{q^{n-2}}\left(\frac{q+np}{q}\right)=\frac{nq+n^{2}p}{q^{n-1}}\because p+q=1$
multiple by $q^{n-1}$
$C_1{q}^{n-1}+2^2{C}_{2}p{q}^{n-2}+3^2{C}_3{p}^2{q}^{n-3}....=nq=n^{2}p$
$C_1{q}^{n-1}+2^2{C}_{2}p{q}^{n-2}+3^2{C}_3{p}^2{q}^{n-3}+....=npq=n^2{p}^2$