Let $$\mathrm{a}_{\mathrm{n}}$$ be the $$\mathrm{n}^{\text {th }}$$ term of the series $$5+8+14+23+35+50+\ldots$$ and $$\mathrm{S}_{\mathrm{n}}=\sum_\limits{k=1}^{n} a_{k}$$. Then $$\mathrm{S}_{30}-a_{40}$$ is equal to :

Let $$S_{K}=\frac{1+2+\ldots+K}{K}$$ and $$\sum_\limits{j=1}^{n} S_{j}^{2}=\frac{n}{A}\left(B n^{2}+C n+D\right)$$, where $$A, B, C, D \in \mathbb{N}$$ and $$A$$ has least value. Then

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 :