Question

${\sum }_{r=0}^n{r}^3.{{}^{n}C}_r{p}^r{q}^{n-r}=$

• A. $=np(p+2q{)}^{n-3}[(p+q{)}^2+(p+q\left)\right(p+1\left)\right(n-1)+p(n-1\left)\right(n-2\left)\right]$
• B. $=np(p+3q{)}^{n-3}[(p+q{)}^2+(p+q\left)\right(p+1\left)\right(n-1)+p(n-1\left)\right(n-2\left)\right]$
• C. $=np(p-4q{)}^{n-3}[(p+q{)}^2+(p+q\left)\right(p+1\left)\right(n-1)+p(n-1\left)\right(n-2\left)\right]$
• D. $=np(p+q{)}^{n-3}[(p+q{)}^2+(p+q\left)\right(p+1\left)\right(n-1)+p(n-1\left)\right(n-2\left)\right]$

Answer 1

Answer: (D)

$=np(p+q{)}^{n-3}[(p+q{)}^2+(p+q\left)\right(p+1\left)\right(n-1)+p(n-1\left)\right(n-2\left)\right]$

$(q+x{)}^n=q^n+{}^n{C}_1{q}^{n-1}{x}^1+{}^n{C}_2{q}^{n-2}{x}^2+{}^n{C}_3{q}^{n-3}{x}^3+...+{}^n{C}_n{q}^0{x}^n$
Differentiating with respect to $x$, we get
$n(q+x{)}^{n-1}={}^n{C}_1{q}^{n-1}{x}^0+2{}^n{C}_2{q}^{n-2}{x}^1+3{}^n{C}_3{q}^{n-3}{x}^2+...+n{}^n{C}_n{q}^0{x}^{n-1}$
Multiplying with $x$ on both sides.
$nx(q+x{)}^{n-1}={}^n{C}_1{q}^{n-1}{x}^1+2{}^n{C}_2{q}^{n-2}{x}^2+3{}^n{C}_3{q}^{n-3}{x}^3+...+n{}^n{C}_n{q}^0{x}^n$
Differentiating with respect to $x$, we get
$n\left[\right(q+x{)}^{n-1}+x(n-1)(q+x{)}^{n-2}]={}^n{C}_1{q}^{n-1}{x}^0+2^2{}^n{C}_2{q}^{n-2}{x}^1+3^2{}^n{C}_3{q}^{n-3}{x}^2+...+n^2{}^n{C}_n{q}^0{x}^{n-1}$
Multiplying by $x$ on both sides, we get
$nx\left[\right(q+x{)}^{n-1}+x(n-1)(q+x{)}^{n-2}]={}^n{C}_1{q}^{n-1}{x}^1+2^2{}^n{C}_2{q}^{n-2}{x}^2+3^2{}^n{C}_3{q}^{n-3}{x}^3+...+n^2{}^n{C}_n{q}^0{x}^n$
Differentiating with respect to $x$, and multiplying with $x$, on both sides we get
$nx\left[\right(q+x{)}^{n-1}+x(n-1)(q+x{)}^{n-2}+(n-1\left)\right(q+x{)}^{n-2}+x(n-1)(n-2)(q+x{)}^{n-3}]=\sum r^3{}^{}{}^n{C}_r{q}^{n-r}{x}^r$
Substituting $x=p$ we get
$\sum r^3{}^n{C}_r{q}^{n-r}{p}^r$
$=np(p+q{)}^{n-3}[(q+p{)}^2+p(n-1\left)\right(q+p)+(n-1\left)\right(q+p)+p(n-1\left)\right(n-2\left)\right]$
$=np(p+q{)}^{n-3}[(p+q{)}^2+(p+q\left)\right(p+1\left)\right(n-1)+p(n-1\left)\right(n-2\left)\right]$