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Let $C_{r}$ denote the coefficient of $x^{r}$ in the binomial expansion of $(1+x)^{n}$, $n \in N$, $0 \leq r \leq n$. If $P_{n} = C_{0} - C_{1} + \frac{2^{2}}{3}C_{2} - \frac{2^{3}}{4}C_{3} + \dots + \frac{(-2)^{n}}{n+1}C_{n}$, then the value of $\sum_{n=1}^{25} \frac{1}{P_{2n}}$ equals.

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

Let $M = 2^{30} - 2^{15} + 1$. When $M^2$ is expressed in base $2$,the number of $1$'s in its binary representation is:

The sum of the coefficients of $x^{499}$ and $x^{500}$ in $(1+x)^{1000}+x(1+x)^{999}+x^{2}(1+x)^{998}+.......+x^{1000}$ is

The sum of the coefficients of all even degree terms in $x$ in the expansion of $(x + \sqrt{x^3 - 1})^6 + (x - \sqrt{x^3 - 1})^6$ for $x > 1$ is equal to:

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