If $(1+x-2x^2)^6 = 1+a_1x+a_2x^2+\ldots+a_{12}x^{12}$, then the value of $a_2+a_4+a_6+\ldots+a_{12}$ is

$3 \cdot C_0 + 7 \cdot C_1 + 11 \cdot C_2 + \ldots + (3 + 4n) C_n =$

Let $\binom{n}{k} = \frac{n!}{k!(n-k)!}$. Then the sum $\frac{1}{2^{10}} \sum_{k=0}^{10} \binom{10}{k} k^2$ lies in the interval

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

The sum of the series $\binom{20}{0} - \binom{20}{1} + \binom{20}{2} - \binom{20}{3} + \dots + \binom{20}{10}$ is:

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