$\sum_{\mathrm{k}=0}^{20}\left({ }^{20} \mathrm{C}_{\mathrm{k}}\right)^{2}$ is equal to :

  • [JEE MAIN 2021]
  • A

    ${ }^{40} \mathrm{C}_{21}$

  • B

    ${ }^{40} \mathrm{C}_{19}$

  • C

    ${ }^{40} \mathrm{C}_{20}$

  • D

    ${ }^{41} \mathrm{C}_{20}$

Similar Questions

If ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_{15}}{x^{15}},$ then ${C_2} + 2{C_3} + 3{C_4} + .... + 14{C_{15}} = $

  • [IIT 1966]

The sum to $(n + 1)$ terms of the following series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + $..... is

If n is a positive integer and ${C_k} = {\,^n}{C_k}$, then the value of ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =

If $\sum_{ r =0}^5 \frac{{ }^{11} C _{2 r +1}}{2 r +2}=\frac{ m }{ n }, \operatorname{gcd}( m , n )=1$, then $m - n$ is equal to _____

  • [JEE MAIN 2025]

If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $, then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

  • [AIEEE 2004]