Question

If ${}^{20}{C}_r$ is the co-efficient of $x^r$ in the expansion of $(1+x{)}^{20},$ then the value $\sum _{r=0}^{20}{r}^2{}^{20}{C}_r$ is equal to:

• A. $420\times 2^{19}$
• B. $420\times 2^{18}$
• C. $380\times 2^{19}$
• D. $380\times 2^{18}$

Answer 1

The correct option is B $420\times 2^{18}$

$\because (1+x{)}^{20}={}^{20}{C}_0+{}^{20}{C}_{1}x+{}^{20}{C}_2{x}^2+\cdots +{}^{20}{C}_{20}{x}^{20}$
On differentiating both sides w.r.t $x$ we get:
$20(1+x{)}^{19}=1\cdot {}^{20}{C}_1+2\cdot {}^{20}{C}_{2}x+\cdots +20\cdot {}^{20}{C}_{20}{x}^{19}$
$\Rightarrow 20(1+x{)}^{19}x=1\cdot {}^{20}{C}_{1}x+2\cdot {}^{20}{C}_2{x}^2+\cdots +20\cdot {}^{20}{C}_{20}{x}^{20}$
Again on differentiating w.r.t. $x$ we get:
$20(1+x{)}^{19}+20\times 19\cdot x(1+x{)}^{18}=1^2\cdot {}^{20}{C}_1+2^2\cdot {}^{20}{C}_{2}x+\cdots +{20}^2\cdot {}^{20}{C}_{20}{x}^{19}$
Replace $x$ by $1,$ we get:
$1^2\cdot {}^{20}{C}_1+2^2\cdot {}^{20}{C}_2+3^2\cdot {}^{20}{C}_3+\cdots +{20}^2\cdot {}^{20}{C}_{20}=20\cdot 2^{19}+20\times 19\cdot 2^{18}$

$\therefore \sum _{r=0}^{20}{r}^2{}^{20}{C}_r=420\times 2^{18}$