If the sum of the first 20 terms of the series $\frac{4 \cdot 1}{4+3 \cdot 1^2+1^4}+\frac{4 \cdot 2}{4+3 \cdot 2^2+2^4}+\frac{4 \cdot 3}{4+3 \cdot 3^2+3^4}+\frac{4 \cdot 4}{4+3 \cdot 4^2+4^4}+\ldots \cdot$ is $\frac{\mathrm{m}}{\mathrm{n}}$, where m and n are coprime, then $\mathrm{m}+\mathrm{n}$ is equal to :

Consider two sets A and B, each containing three numbers in A.P. Let the sum and the product of the elements of A be 36 and p respectively and the sum and the product of the elements of B be 36 and $q$ respectively. Let d and D be the common differences of $\mathrm{AP}^{\prime} \mathrm{s}$ in $A$ and $B$ respectively such that $D=d+3, d>0$. If $\frac{p+q}{p-q}=\frac{19}{5}$, then $\mathrm{p}-\mathrm{q}$ is equal to

Let $A=\{1,6,11,16, \ldots\}$ and $B=\{9,16,23,30, \ldots\}$ be the sets consisting of the first 2025 terms of two arithmetic progressions. Then $n(A \cup B)$ is

$1+3+5^2+7+9^2+\ldots$ upto 40 terms is equal to