Question

If $C_r$ be the coefficient of $x^r$ in ${\left(1+x\right)}^n$ , then the value of
$\sum _{r=0}^n{\left(r+1\right)}^2{C}_r$ is :

• A. $\left(n+1\right)\left(n+4\right)2^{n-2}$
• B. $\left(n+1\right)\left(n+4\right)2^{n-1}$
• C. ${\left(n+1\right)}^2{.2}^{n-2}$
• D. ${\left(n+4\right)}^2{.2}^{n-2}$

Answer 1

The correct option is A $\left(n+1\right)\left(n+4\right)2^{n-2}$

$(1+x{)}^n=\sum C_r{x}^r.....\left(1\right)$

$n(1+x{)}^{n-1}=\sum r{C}_r{x}^{r-1}......\left(2\right)$

$n(n-1)(1+x{)}^{n-2}=\sum r(r-1)C_r{x}^{r-2}....\left(3\right)$

Put $x=1$ in $\left(1\right),\left(2\right),\left(3\right)$

$\sum C_r=2^n,\sum r{C}_r=n{.2}^{n-1},\sum (r^2-r)C_r=n(n-1)2^{n-2}⟹\sum r^2{C}_r=(n^2-n)2^{n-2}+2n{.2}^{n-2}=(n^2+n){.2}^{n-2}$

$\sum (r+1{)}^2{C}_r=\sum r^2{C}_r+2\sum r{C}_r+\sum C_r=(n^2+n)2^{n-2}+4n{.2}^{n-2}+{4.2}^{n-2}=(n^2+5n+4)2^{n-2}=(n+1\left)\right(n+4){.2}^{n-2}$