$\frac{{{C_1}}}{{{C_0}}} + 2\frac{{{C_2}}}{{{C_1}}} + 3\frac{{{C_3}}}{{{C_2}}} + .... + 15\frac{{{C_{15}}}}{{{C_{14}}}} = $

  • [IIT 1962]
  • A

    $100$

  • B

    $120$

  • C

    $- 120$

  • D

    એકપણ નહીં.

Similar Questions

જો ${(1 + x)^n} = {C_0} + {C_1}x + {C_2}{x^2} + .......... + {C_n}{x^n}$, તો $\frac{{{C_1}}}{{{C_0}}} + \frac{{2{C_2}}}{{{C_1}}} + \frac{{3{C_3}}}{{{C_2}}} + .... + \frac{{n{C_n}}}{{{C_{n - 1}}}} = $

$(\alpha + p)^{m - 1} + (\alpha + p)^{m - 2} (\alpha + q) + (\alpha + p)^{m - 3} (\alpha + q)^2 + ...... (\alpha + q)^{m - 1}$ 

વિસ્તરણમાં $\alpha ^t$ નો સહગુણક મેળવો.

જ્યાં $\alpha \ne - q$ અને $p \ne q$  

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = $

  • [JEE MAIN 2017]

${(1 + x)^{50}}$ ના વિસ્તરણમાં $x$ ની અયુગ્મ ઘાતાંકના સહગુણકનો સરવાળો મેળવો.

જો $\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}$ હોય,તો $n=..........$

  • [JEE MAIN 2023]