The sum of n terms of m A.P.s are S₁, S₂, S₃, …, S_m. If  the first term and common difference are 1, 2, 3, …, m  respectively, then S₁ + S₂ + S₃ + … + $S_m$ =

We have, $\begin{array}{l}{\mathrm{S}}_1=(\mathrm{n}/2)[2\cdot 1+(\mathrm{n}-1)\cdot 1]\\ {\mathrm{S}}_2=(\mathrm{n}/2)[2\cdot 2+(\mathrm{n}-1)\cdot 2]\\ {\mathrm{S}}_{\mathrm{m}}=(\mathrm{n}/2)[2\cdot \mathrm{m}+(\mathrm{n}-1)\cdot \mathrm{m}]\end{array}$

$\begin{array}{l}\therefore {\mathrm{S}}_1+{\mathrm{S}}_2+\dots +{\mathrm{S}}_{\mathrm{m}}\\ =\mathrm{n}(1+2+3\dots +\mathrm{m})+\frac{\mathrm{n}(\mathrm{n}-1)}{2}\times (1+2+\dots +\mathrm{m})\\ =\frac{\mathrm{m}(\mathrm{m}+1)}{2}\left(\mathrm{n}+\frac{{\mathrm{n}}^2-\mathrm{n}}{2}\right)\\ =\frac{\mathrm{m}(\mathrm{m}+1)}{2}\cdot \frac{\mathrm{n}(\mathrm{n}+1)}{2}=\frac{1}{4}\mathrm{mn}(\mathrm{m}+1)(\mathrm{n}+1)\end{array}$