If $(1 + x) (1 + x + x^2) (1 + x + x^2 + x^3) \dots (1 + x + x^2 + \dots + x^n) \equiv a_0 + a_1x + a_2x^2 + a_3x^3 + \dots + a_mx^m$,then $\sum_{r=0}^m a_r$ has the value equal to:

Let $a > b > 0$ and $f(n) = a^{1/n} - b^{1/n}$, $J(n) = (a - b)^{1/n}$ for all $n \geq 2$. Then:

Let $K$ be the number of rational terms in the expansion of $(\sqrt{2}+\sqrt[3]{3})^{6144}$. If the coefficient of $x^{P} \quad(P \in N)$ in the expansion of $\frac{1}{(1+x)(1+x^2)(1+x^4)(1+x^8)(1+x^{16})}$ is $\alpha_{P}$,then $\alpha_{K}-\alpha_{K+1}-\alpha_{K-1}=$

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-

If the $6^{th}$ term in the expansion of the binomial $[\sqrt{2^{\log(10 - 3^x)}} + \sqrt[5]{2^{(x - 2)\log 3}}]^m$ is equal to $21$ and it is known that the binomial coefficients of the $2^{nd}$,$3^{rd}$ and $4^{th}$ terms in the expansion represent respectively the first,third and fifth terms of an $A.P.$ (the symbol $\log$ stands for logarithm to the base $10$),then $x = $

If $C_r$ denotes the binomial coefficient ${ }^{n} C_r$,then $(-1) C_0^2+2 C_1^2+5 C_2^2+\ldots+(3 n-1) C_n^2$ is equal to