Let $\mathrm{S}=\frac{1}{25!}+\frac{1}{3!23!}+\frac{1}{5!21!}+\ldots$ up to 13 terms. If $13 \mathrm{~S}=\frac{2^k}{n!}, k \in \mathrm{~N}$, then $n+k$ is equal to

The sum of all possible values of $\mathbf{n} \in \mathbf{N}$, so that the coefficients of $x, x^2$ and $x^3$ in the expansion of $\left(1+x^2\right)^2(1+x)^{\mathrm{n}}$, are in arithmetic progression is :

The value of $\frac{{ }^{100} \mathrm{C}_{50}}{51}+\frac{{ }^{100} \mathrm{C}_{51}}{52}+\ldots .+\frac{{ }^{100} \mathrm{C}_{100}}{101}$ is:

Let $\mathrm{C}_{\mathrm{r}}$ denote the coefficient of $x^{\mathrm{r}}$ in the binomial expansion of $(1+x)^{\mathrm{n}}, \mathrm{n} \in \mathrm{N}, 0 \leq \mathrm{r} \leq \mathrm{n}$. If

$P_n=C_0-C_1+\frac{2^2}{3} C_2-\frac{2^3}{4} C_3+\ldots . .+\frac{(-2)^n}{n+1} C_n$, then the value of $\sum\limits_{n=1}^{25} \frac{1}{P_{2 n}}$ equals.