If $$\operatorname{gcd}~(\mathrm{m}, \mathrm{n})=1$$ and $$1^{2}-2^{2}+3^{2}-4^{2}+\ldots . .+(2021)^{2}-(2022)^{2}+(2023)^{2}=1012 ~m^{2} n$$ then $$m^{2}-n^{2}$$ is equal to :

The sum of the first $$20$$ terms of the series $$5+11+19+29+41+\ldots$$ is :

The sum $$\sum\limits_{n = 1}^\infty {{{2{n^2} + 3n + 4} \over {(2n)!}}} $$ is equal to :

The sum of 10 terms of the series

$${1 \over {1 + {1^2} + {1^4}}} + {2 \over {1 + {2^2} + {2^4}}} + {3 \over {1 + {3^2} + {3^4}}}\, + \,....$$ is