Question

${C_0}^2+\frac{1}{2}{C_1}^2+\frac{1}{3}{C_2}^2+....\frac{1}{n+1}{C_n}^2$ equals

• A. $\frac{1}{n}\left({{}^{2n}C}_n\right)$
• B. $\frac{1}{n+1}\left({{}^{2n+1}C}_n\right)$
• C. $\frac{1}{n}\left({{}^{2n}C}_{n+1}\right)$
• D. none of these

Answer 1

The correct option is B $\frac{1}{n+1}\left({{}^{2n+1}C}_n\right)$

Consider,
$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+.....+C_n{x}^n$
Integrating both sides w.r.t. $x$, within the limits $0$ to $x$, we get
${\int }_0^x(1+x{)}^{n}dx={\int }_0^x(C_0+C_{1}x+C_2{x}^2+C_3{x}^3+.....+C_n{x}^n)dx$
$\Rightarrow \frac{(1+x{)}^{n+1}-1}{1+n}=C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+.....+\frac{C_n{x}^{n+1}}{n+1}$ ....(1)
Now, $(x+1{)}^n=C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+.....+C_n$ .....(2)
Multiplying (1) and (2), we get
$\Rightarrow \frac{(1+x{)}^{2n+1}-(1+x{)}^n}{1+n}=(C_{0}x+\frac{C_1{x}^2}{2}+\frac{C_2{x}^3}{3}+.....+\frac{C_n{x}^{n+1}}{n+1})(C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+.....+C_n)$ .....(3)
Coefficient of $x^{n+1}$ in LHS of (3) is $\frac{{}^{2n+1}{C}_n}{n+1}$.
And coefficient of $x^{n+1}$ in RHS of (3) is $C_0^2+\frac{C_1^2}{2}+\frac{C_2^2}{3}+.....+\frac{C_n^2}{n+1}$.
Since, eqn (3) is an identity,
$\Rightarrow$coefficient of $x^{n+1}$ in LHS of (3)=coefficient of $x^{n+1}$ in RHS of (3)
$\Rightarrow \frac{{}^{2n+1}{C}_n}{n+1}=C_0^2+\frac{C_1^2}{2}+\frac{C_2^2}{3}+.....+\frac{C_n^2}{n+1}$