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} = $

Let $S_n = 1 + q + q^2 + ..... + q^n$ and $T_n = 1 + \left( \frac{q + 1}{2} \right) + \left( \frac{q + 1}{2} \right)^2 + ...... + \left( \frac{q + 1}{2} \right)^n$ where $q$ is a real number and $q \ne 1$. If $^{101}C_1 + ^{101}C_2 \cdot S_1 + ...... + ^{101}C_{101} \cdot S_{100} = \alpha \cdot T_{100}$,then $\alpha$ is equal to

Let the coefficient of $x^{r}$ in the expansion of $(x+3)^{n-1}+(x+3)^{n-2}(x+2)+(x+3)^{n-3}(x+2)^2+\ldots+(x+2)^{n-1}$ be $\alpha_{r}$. If $\sum_{r=0}^{n-1} \alpha_{r}=\beta^{n}-\gamma^{n}$,where $\beta, \gamma \in N$,then the value of $\beta^2+\gamma^2$ equals:

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 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 = $