The sum of the series $\sum\limits_{r = 0}^n {{{( - 1)}^r}\,{\,^n}{C_r}\left( {\frac{1}{{{2^r}}} + \frac{{{3^r}}}{{{2^{2r}}}} + \frac{{{7^r}}}{{{2^{3r}}}} + \frac{{{{15}^r}}}{{{2^{4r}}}} + .....m\,{\rm{terms}}} \right)} $ is
$\frac{{{2^{mn}} - 1}}{{{2^{mn}}({2^n} - 1)}}$
$\frac{{{2^{mn}} - 1}}{{{2^n} - 1}}$
$\frac{{{2^{mn}} + 1}}{{{2^n} + 1}}$
None of these
The sum to $(n + 1)$ terms of the series $\frac{{{C_0}}}{2} - \frac{{{C_1}}}{3} + \frac{{{C_2}}}{4} - \frac{{{C_3}}}{5} + ...$ is
If ${C_0},{C_1},{C_2},.......,{C_n}$ are the binomial coefficients, then $2.{C_1} + {2^3}.{C_3} + {2^5}.{C_5} + ....$ equals
For $x\, \in \,R\,,\,x\, \ne \, - 1,$ if ${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ....{x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i\,}{\,x^i}} ,$ then $a_{17}$ is equal to
Let $(1+2 x)^{20}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$.Then $3 a_0+2 a_1+3 a_2+2 a_3+3 a_4+2 a_5+\ldots+2 a_{19}+3 a_{20}$ equals
The value of $\frac{{{C_1}}}{2} + \frac{{{C_3}}}{4} + \frac{{{C_5}}}{6} + .....$ is equal to