$\sum\limits_{k = 0}^{10} {^{20}{C_k} = }$

$\frac{C_1}{C_0} + 2\frac{C_2}{C_1} + 3\frac{C_3}{C_2} + \dots + 15\frac{C_{15}}{C_{14}} = $

For $r=0, 1, \ldots, 10$,let $A_{r}, B_{r}$ and $C_{r}$ denote,respectively,the coefficient of $x^{r}$ in the expansions of $(1+x)^{10}$,$(1+x)^{20}$ and $(1+x)^{30}$. Then $\sum_{r=1}^{10} A_r(B_{10} B_r - C_{10} A_r)$ is equal to

If $(\frac{1}{^{15}C_{0}}+\frac{1}{^{15}C_{1}})(\frac{1}{^{15}C_{1}}+\frac{1}{^{15}C_{2}})...(\frac{1}{^{15}C_{12}}+\frac{1}{^{15}C_{13}}) = \frac{a^{13}}{^{14}C_{0} \cdot ^{14}C_{1} \cdot ... \cdot ^{14}C_{12}}$, then $30a$ is equal to:

If $(1 + x)^{15} = C_0 + C_1x + C_2x^2 + ...... + C_{15}x^{15},$ then $C_2 + 2C_3 + 3C_4 + .... + 14C_{15} = $

The sum of the last $30$ coefficients in the expansion of $(1+x)^{59}$, when expanded in ascending powers of $x$, is: