If $T_r = ^{2016}C_r x^{2016-r}$ for $r = 0, 1, 2, \dots, 2016$,then $(T_0 - T_2 + T_4 - \dots + T_{2016})^2 + (T_1 - T_3 + T_5 - \dots - T_{2015})^2$ is equal to-

Let $C_{r}$ denote the binomial coefficient of $x^{r}$ in the expansion of $(1+x)^{10}$. If $\alpha, \beta \in R$,and $C_{1}+3 \cdot 2 C_{2}+5 \cdot 3 C_{3}+\ldots$ (up to $10$ terms) $= \frac{\alpha \times 2^{11}}{2^{\beta}-1} \left( C_{0}+\frac{C_{1}}{2}+\frac{C_{2}}{3}+\ldots \right.$ (up to $10$ terms) $)$,then the value of $\alpha+\beta$ is equal to:

If $a_r$ is the coefficient of $x^r$ in the expansion of $(1 + x + x^2)^n$,then $a_1 - 2a_2 + 3a_3 - \dots - 2n\,a_{2n} = $

The value of $\sum\limits_{n = 0}^4 {{{\left( {1009 - 2n} \right)}^4} \binom{4}{n} {\left( { - 1} \right)^n}}$ is

Let $\alpha = \sum_{r=0}^{n} (4r^2+2r+1) {}^{n}C_{r}$ and $\beta = \left(\sum_{r=0}^{n} \frac{{}^{n}C_{r}}{r+1}\right) + \frac{1}{n+1}$. If $140 < \frac{2\alpha}{\beta} < 281$,then the value of $n$ is...............

Let $R=(5 \sqrt{5}+11)^{2 n+1}$ and $f=R-[R]$,where $[x]$ denotes the greatest integer less than or equal to $x$,then $R f=$