Question

The value of $\sum _{r=0}^n(2r+1)({}^n{C}_r{)}^2$ is equal to :

• A. (2n+2). ²ⁿC_n
• B. (n+1). ²ⁿC_n
• C. (2n+1). ²ⁿC_n
• D. n. ²ⁿC_n

Answer 1

The correct option is B

(n+1). ²ⁿC_n

$\sum _{r=0}^n(2r+1)({}^n{C}_r{)}^2$

= 2 $\sum _0^{n}r{C}_r^2+\sum _0^{n}r{C}_r^2+2A+B$

$C_0+C_1+C_2{x}^2+...+C_n{x}^n=(1+x{)}^n$

Differentiatingbg w.r.t x , we get

$C_1+2C_2+3C_3{x}^2+....+n{x}^{n-1}=n(1+x{)}^{n-1}$

Also $C_0{x}^n+C_1{x}^{n-1}+C_2{x}^{n-2}+.....+C_n=(x+1{)}^n$

$\therefore$ $C_1^2+2C_3^2+3C_3^2+...+n{C}_n^2$

= coeff. of $x^{n-1}inn(1+x{)}^{2n-1}$

= n. ${}^{2n-1}{C}_{n-1}$

Also $C_0^2+C_1^2+C_2^2+...+C_n^2=2^n{C}_n$

$\therefore$ 2A + B = 2n. ${}^{2n-1}{C}_{n-1}+{}^{2n}{C}_n$

= 2n. $\frac{(2n-1)!}{(n-1)!n!}+\frac{2n!}{n!n!}=n.{}^{2n}{C}_n+{}^{2n}{C}_n$

= (n+1) ${}^{2n}{C}_n$

Alter: S = $\sum _{r=0}^{n}2(n-r)+1({}^n{C}_r{)}^2$

= $\sum _{r=0}^{n}2(n-r)+1({}^n{C}_{n-r}{)}^2$