The value of $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ is equal to
$\frac{{{2^n} - 1}}{{n + 1}}$
$n{.2^n}$
$\frac{{{2^n}}}{n}$
$\frac{{{2^n} + 1}}{{n + 1}}$
The sum, of the coefficients of the first $50$ terms in the binomial expansion of $(1-x)^{100}$, is equal to
The value of $\sum_{ r =0}^{6}\left({ }^{6} C _{ r }{ }^{-6} C _{6- r }\right)$ is equal to :
$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to
Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ then $n + m$ is equal to (Here $\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)$
If n is a positive integer and ${C_k} = {\,^n}{C_k}$, then the value of ${\sum\limits_{k = 1}^n {{k^3}\left( {\frac{{{C_k}}}{{{C_{k - 1}}}}} \right)} ^2}$ =