Question

The value of the sum ${\left({}^n{C}_1\right)}^2+{\left({}^n{C}_2\right)}^2+{\left({}^n{C}_3\right)}^2+\cdots +{\left({}^n{C}_n\right)}^2$ is

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

Answer 1

The correct option is D ${}^{2n}{C}_n-1$

We know that

${(1+x)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+\cdots +{}^n{C}_n{x}^n$ .....(i)
and

${(x+1)}^n={}^n{C}_0{x}^n+{}^n{C}_1{x}^{n-1}+{}^n{C}_2{x}^{n-2}+\cdots +{}^n{C}_n$ .....(ii)

On multiplying equations (i) and (ii), we get

${(1+x)}^{2n}=({}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+\cdots {+}^n{C}_n{x}^n)\times ({}^n{C}_0{x}^n{+}^n{C}_1{x}^{n-1}{+}^n{C}_2{x}^{n-2}+\cdots {+}^n{C}_n)$

Coefficient of $x^n$ in right hand side $={\left({}^n{C}_0\right)}^2+{\left({}^n{C}_1\right)}^2+\cdots +{\left({}^n{C}_n\right)}^2$

and

coefficient of $x^n$ in left hand side $={}^{2n}{C}_n$

$\therefore {\left({}^n{C}_0\right)}^2+{\left({}^n{C}_1\right)}^2+\cdots +{\left({}^n{C}_n\right)}^2=\frac{2n!}{n!n!}$

$\Rightarrow {\left({}^n{C}_1\right)}^2+\cdots +{\left({}^n{C}_n\right)}^2=\frac{\left(2n\right)!}{n!n!}-1$ $={}^{2n}{C}_n-1$