The remainder when the polynomial $1+x^2+x^4+x^6+\ldots+x^{22}$ is divided by $1+x+x^2+x^3+\ldots+x^{11}$ is

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 $\alpha_{K}-\alpha_{K+1}-\alpha_{K-1}=$

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 $(1+x+x^2)^n = c_0 + c_1 x + c_2 x^2 + \ldots$,then the value of $c_0 c_1 - c_1 c_2 + c_2 c_3 - \ldots$ 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 $(1 + x - 3x^2)^{2145} = a_0 + a_1x + a_2x^2 + \dots$,then $a_0 - a_1 + a_2 - a_3 + \dots$ ends with: