$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ =  

  • [JEE MAIN 2019]
  • A

    ${2^{23}}$

  • B

    ${2^{26}}$

  • C

    ${2^{24}}$

  • D

    ${2^{25}}$

Similar Questions

$\sum\limits_{n = 0}^4 {{{\left( {1009 - 2n} \right)}^4}\left( \begin{gathered}
  4 \hfill \\
  n \hfill \\ 
\end{gathered}  \right)} {\left( { - 1} \right)^n}$   ની કિમત મેળવો 

$n\left[ {x - \left( {\frac{{^n{C_0}{ + ^n}{C_1}}}{{^n{C_0}}}} \right)} \right]\left[ {\frac{x}{2} - \left( {\frac{{^n{C_1}{ + ^n}{C_2}}}{{^n{C_1}}}} \right)} \right]\left[ {\frac{x}{3} - \left( {\frac{{^n{C_2}{ + ^n}{C_3}}}{{^n{C_2}}}} \right)} \right].....$ $ \left[ {\frac{x}{n} - \left( {\frac{{^n{C_{n - 1}}{ + ^n}{C_n}}}{{^n{C_{n - 1}}}}} \right)} \right]$ ના વિસ્તરણમાં $x^{n-6}$ નો સહગુણક મેળવો 

(જ્યાં $n = n . (n -1) . (n -2).... 3.2.1$)

જો ${C_r}$ એ $^n{C_r}$ દર્શાવે છે તો , $\frac{{2(n/2)!(n/2)!}}{{n!}}[C_0^2 - 2C_1^2 + 3C_2^2 - ..... + {( - 1)^n}(n + 1)C_n^2]$ મેળવો. (કે જ્યાં $n$ એ યુગ્મ પુર્ણાક છે )

  • [IIT 1986]

${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ = . . .

${C_0}{C_r} + {C_1}{C_{r + 1}} + {C_2}{C_{r + 2}} + .... + {C_{n - r}}{C_n}$=