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 the coefficient of $x$ in the expansion of $(ax^{2}+bx+c)(1-2x)^{26}$ is $-56$ and the coefficients of $x^{2}$ and $x^{3}$ are both zero,then $a+b+c$ is equal to:

Let $\lambda$ be the positive root of the equation $x^2-x-1=0$,and set $a_n = \frac{1}{\sqrt{5}}\left(\lambda^n - (1-\lambda)^n\right)$ for $n \in N$,where $N$ is the set of all natural numbers. Consider the sets $A = \{ n \in N : a_n \text{ is a rational number, but not an integer} \}$ and $B = \{ n \in N : a_n \text{ is an irrational number} \}$. Then:

If $C_0, C_1, C_2, \ldots, C_{n}$ are the binomial coefficients in the expansion of $(1+x)^{n}$,then $(C_0+C_1)-(C_2+C_3)+(C_4+C_5)-(C_6+C_7)+\ldots=$

For an integer $n \geq 2$,if the arithmetic mean of all coefficients in the binomial expansion of $(x+y)^{2n-3}$ is $16$,then the distance of the point $P(2n-1, n^2-4n)$ from the line $x+y=8$ is: