The coefficient of $t^{24}$ in the expansion of $(1 + t^2)^{12}(1 + t^{12})(1 + t^{24})$ is

If $(1 - x + 2x^2)^n = a_0 + a_1x + a_2x^2 + \dots + a_{2n}x^{2n}$,where $n \in N$,$x \in R$,and $a_0, a_1, a_2$ are in Arithmetic Progression $(A.P.)$,then there exists:

If $1 + (2 + {}^{49}C_{1} + {}^{49}C_{2} + \dots + {}^{49}C_{49})({}^{50}C_{2} + {}^{50}C_{4} + \dots + {}^{50}C_{50})$ is equal to $2^{n} \cdot m$,where $m$ is odd,then $n + m$ is equal to.

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-

The sum of the series $\sum\limits_{r = 0}^n {(-1)^r \, ^nC_r \left( \frac{1}{2^r} + \frac{3^r}{2^{2r}} + \frac{7^r}{2^{3r}} + \frac{15^r}{2^{4r}} + \dots + m \text{ terms} \right)}$ is

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: