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 sum of the coefficients of three consecutive terms in the binomial expansion of $(1+ x )^{ n +2}$, which are in the ratio $1: 3: 5$, is equal to

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

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

The value of ${\sum\limits_{r = 1}^{19} {\frac{{{}^{20}{C_{r + 1}}\left( { - 1} \right)}}{{{2^{2r + 1}}}}} ^r}$ is

Let ${\left( {1 + x + {x^2}} \right)^{20}}\left( {2x + 1} \right) = {a_0} + {a_1}{x^1} + {a_2}{x^2} + ... + {a_{41}}{x^{41}}$ , then $\frac{{{a_0}}}{1} + \frac{{{a_1}}}{2} + .... + \frac{{{a_{41}}}}{{42}}$ is equal to