If $C_{x} \equiv^{25} C_{x}$ and $\mathrm{C}_{0}+5 \cdot \mathrm{C}_{1}+9 \cdot \mathrm{C}_{2}+\ldots .+(101) \cdot \mathrm{C}_{25}=2^{25} \cdot \mathrm{k}$ then $\mathrm{k}$ is equal to

Find the coefficient of $x^{49}$ in the expansion of $(2x + 1) (2x + 3) (2x + 5)----- (2x + 99)$

The sum of the last eight coefficients in the expansion of ${(1 + x)^{15}}$ is

${n^n}{\left( {\frac{{n + 1}}{2}} \right)^{2n}}$ is

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

Let $K$ be the sum of the coefficients of the odd powers of $x$ in the expansion of $(1+ x )^{99}$. Let a be the middle term in the expansion of $\left(2+\frac{1}{\sqrt{2}}\right)^{200}$. If $\frac{{ }^{200} C _{99} K }{ a }=\frac{2^{\ell} m }{ n }$, where $m$ and $n$ are odd numbers, then the ordered pair $(l, n )$ is equal to :