Question

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

• A. $\frac{(n+2)(2n-1)!}{n!(n-1)!}$
• B. $\frac{(n+2)(2n+1)!}{n!(n-1)!}$
• C. $\frac{(n+2)(2n+1)!}{n!(n+1)!}$
• D. $\frac{(n+2)(2n-1)!}{n!(n+1)!}$

Answer 1

The correct option is A $\frac{(n+2)(2n-1)!}{n!(n-1)!}$

Given

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

It can be written as

${\sum }_{r=0}^{n}r\cdot C_r^2+{\sum }_{r=0}^n{C}_r^2$

We know that

$r\cdot {}^n{C}_r=n\cdot {}^{n-1}{C}_{r-1}$

Hence

$r\cdot C_r^2=n\cdot {}^{n-1}{C}_{r-1}^2$

${\sum }_{r=0}^{n}n\cdot {}^{n-1}{C}_{r-1}^2=n[{}^{n-1}{C}_0^2+{}^{n-1}{C}_1^2+.......+{}^{n-1}{C}_{n-1}^2]=n\cdot \frac{(2n-1)!}{(n-1)!(2n-1-n+1)!}=n\cdot \frac{(2n-1)!}{(n-1)!n!}----(1)$

${\sum }_{r=0}^n{C}_r^2=\frac{2n!}{n!n!}-----\left(2\right)$

Adding both eq(1) and (2)

${\sum }_{r=0}^n(r+1)C_r^2=n\cdot \frac{(2n-1)!}{(n-1)!n!}+\frac{2n!}{n!n!}$

${\sum }_{r=0}^n(r+1)C_r^2=n\cdot \frac{(2n-1)!}{(n-1)!n!}+\frac{2n(2n-1)!}{n(n-1)!n!}$

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