Let $f(x)=x^{2025}-x^{2000}, x \in[0,1]$ and the minimum value of the function $f(x)$ in the interval $[0,1]$ be $(80)^{80}(n)^{-81}$. Then $n$ is equal to

If the sum of the first four terms of an A.P. is 6 and the sum of its first six terms is 4 , then the sum of its first twelve terms is

If a random variable $x$ has the probability distribution

$$ \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \hline \mathrm{P}(x) & 0 & 2 \mathrm{k} & \mathrm{k} & 3 \mathrm{k} & 2 \mathrm{k}^2 & 2 \mathrm{k} & \mathrm{k}^2+\mathrm{k} & 7 \mathrm{k}^2 \\ \hline \end{array} $$

$$ \text { then } \mathrm{P}(3 < x \leq 6) \text { is equal to } $$

The coefficient of $x^{48}$ in $(1+x)+2(1+x)^2+3(1+x)^3+\ldots+100(1+x)^{100}$ is equal to