Question

$C_0^2+\frac{C_1^2}{2}+\frac{C_2^2}{3}+\dots \dots +\frac{C_n^2}{n+1}=\frac{(2n+1)!}{\left[\right(n+1)!{)}^2}$

• A. True
• B. False

Answer 1

The correct option is A True

${\sum }_{r=0}^m\frac{n{c}_r^m{c}_r}{a+1}=$ ${\sum }_{r=0}^m\frac{n_r{r}^n{C}_r(n+1)}{(n+1)(n+1)}$

${\sum }_{r=0}^m\frac{m_{(n}^{n+1}{C}_{n+1}}{n+1}$ ${}^n{C}_r=\frac{n}{r}{}^{n-1}{C}_{r+1}$

$\frac{2n+1C_{m+1}}{n+1}\Rightarrow \frac{(2n+1)!}{m!(n+1)!(n+1)}$

$\frac{(2n+1)!}{\left[\right(n+1)!{]}^2}$