The sum of the coefficients in the expansion of ${(1 + x - 3{x^2})^{2163}}$ will be

$\left( {\begin{array}{*{20}{c}}n\\0\end{array}} \right) + 2\,\left( {\begin{array}{*{20}{c}}n\\1\end{array}} \right) + {2^2}\left( {\begin{array}{*{20}{c}}n\\2\end{array}} \right) + ..... + {2^n}\left( {\begin{array}{*{20}{c}}n\\n\end{array}} \right)$ is equal to

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

The expression $x^3 - 3x^2 - 9x + c$ can be written in the form $(x - a)^2 (x - b)$ if the values of $c$ is

If $\sum\limits_{ k =1}^{31}\left({ }^{31} C _{ k }\right)\left({ }^{31} C _{ k -1}\right)-\sum\limits_{ k =1}^{30}\left({ }^{30} C _{ k }\right)\left({ }^{30} C _{ k -1}\right)=\frac{\alpha(60 !)}{(30 !)(31 !)}$

Where $\alpha \in R$, then the value of $16 \alpha$ is equal to

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