If ${S_n} = \sum\limits_{r = 0}^n {\frac{1}{{^n{C_r}}}} $ and ${t_n} = \sum\limits_{r = 0}^n {\frac{r}{{^n{C_r}}}} $, then $\frac{{{t_n}}}{{{S_n}}}$ is equal to

If ${a_k} = \frac{1}{{k(k + 1)}},$ for $k = 1,\,2,\,3,\,4,.....,\,n$, then ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

$2.{}^{20}{C_0} + 5.{}^{20}{C_1} + 8.{}^{20}{C_2} + 11.{}^{20}{C_3} + ......62.{}^{20}{C_{20}}$ is equal to

$^{10}{C_1}{ + ^{10}}{C_3}{ + ^{10}}{C_5}{ + ^{10}}{C_7}{ + ^{10}}{C_9} = $

Let ${\left( {1 + x} \right)^{10}} = \sum\limits_{r = 0}^{10} {{C_r}{x^r}} $ and ${\left( {1 + x} \right)^7} = \sum\limits_{r = 0}^7 {{d_r}{x^r}} $ . If $P = \sum\limits_{r = 0}^5 {{C_{2r}}} $ and $Q = \sum\limits_{r = 0}^3 {{d_{2r + 1}}} $ , then $\frac{P}{{2Q}}$ is equal to

The coefficient of $x^{91}$ in the series $^{100}{C_1}\,{2^8}.\,{\left( {1\, - \,x} \right)^{99}}\, + {\,^{100}}{C_2}\,{2^7}.\,{\left( {1\, - \,x} \right)^{98}}\, + {\,^{100}}{C_3}\,{2^6}.\,{\left( {1\, - \,x} \right)^{97}}\, + \,....\, + {\,^{100}}{C_9}\,{\left( {1\, - \,x} \right)^{91}}$ is equal to -