If $\frac{1}{n+1}{ }^n C_n+\frac{1}{n}{ }^n C_{n-1}+\ldots+\frac{1}{2}{ }^{ n } C _1+{ }^{ n } C _0=\frac{1023}{10}$ then $n$ is equal to
$6$
$9$
$8$
$7$
If $n$ be a positive integer such that $n \ge 3$, then the value of the sum to $n$ terms of the series $1 . n - \frac{{\left( {n\, - \,1} \right)}}{{1\,\,!}} (n - 1) + \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)}}{{2\,\,!}} (n - 2) $$- \frac{{\left( {n\, - \,1} \right)\,\,\left( {n\, - \,2} \right)\,\,\left( {n\, - \,3} \right)}}{{3\,\,!}} (n - 3) + ......$ is
If ${}^{21}{C_1} + 3.{}^{21}{C_3} + 5.{}^{21}{C_5} + ......19{}^{21}{C_{19}} + 21.{}^{21}{C_{21}} = k$ Then number of prime factors of $k$ is
The value of ${\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}\left( { - 1} \right)}}{{{2^{2r + 1}}}}} ^r}$ is
If $a$ and $d$ are two complex numbers, then the sum to $(n + 1)$ terms of the following series $a{C_0} - (a + d){C_1} + (a + 2d){C_2} - ........$ is
The sum of the series $\left( {\begin{array}{*{20}{c}}{20}\\0\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{20}\\1\end{array}} \right)$$+$$\left( {\begin{array}{*{20}{c}}{20}\\2\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{20}\\3\end{array}} \right)$$+…..-……+$$\left( {\begin{array}{*{20}{c}}{20}\\{10}\end{array}} \right)$