Suppose $2-p, p, 2-\alpha, \alpha$ are the coefficients of four consecutive terms in the expansion of $(1+x)^n$. Then the value of $p^2-\alpha^2+6\alpha+2p$ equals

The sum of the coefficients of even powers of $x$ in the expansion of ${(1 + x + x^2 + x^3)^5}$ 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 the sum of the coefficients of all even powers of $x$ in the product $(1+x+x^{2}+\ldots+x^{2n})(1-x+x^{2}-x^{3}+\ldots+x^{2n})$ is $61$,then $n$ is equal to

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