If $\frac{{ }^{11} C_1}{2}+\frac{{ }^{11} C_2}{3}+\ldots . .+\frac{{ }^{11} C_9}{10}=\frac{n}{m}$ with $\operatorname{gcd}(n, m)=1$, then $n+m$ is equal to

If ${(1 + x)^{15}} = {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_{15}}{x^{15}},$ then ${C_2} + 2{C_3} + 3{C_4} + .... + 14{C_{15}} = $

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

If $^{20}{C_1} + \left( {{2^2}} \right){\,^{20}}{C_3} + \left( {{3^2}} \right){\,^{20}}{C_3} + \left( {{2^2}} \right) + ..... + \left( {{{20}^2}} \right){\,^{20}}{C_{20}} = A\left( {{2^\beta }} \right)$, then the ordered pair $(A, \beta )$ is equal to

$\mathrm{b}=1+\frac{{ }^1 \mathrm{C}_0+{ }^1 \mathrm{C}_1}{1 !}+\frac{{ }^2 \mathrm{C}_0+{ }^2 \mathrm{C}_1+{ }^2 \mathrm{C}_2}{2 !}+\frac{{ }^3 \mathrm{C}_0+{ }^3 \mathrm{C}_1+{ }^3 \mathrm{C}_2+{ }^3 \mathrm{C}_3}{3 !}+\ldots$

Let $\mathrm{a}=1+\frac{{ }^2 \mathrm{C}_2}{3 !}+\frac{{ }^3 \mathrm{C}_2}{4 !}+\frac{{ }^4 \mathrm{C}_2}{5 !}+\ldots$, Then $\frac{2 b}{a^2}$ is equal to.........................

The value of $^{15}C_0^2{ - ^{15}}C_1^2{ + ^{15}}C_2^2 - ....{ - ^{15}}C_{15}^2$ is