Question

The sum of n terms of m arithmetical progressions are $S_1,S_2,S_3,$_______$S_m$. The first term and common differences are $1,2,3$,_____$m$ respectively. Prove that $S_1+S_2+S_3+$_______$+S_m=\frac{1}{4}mn(m+1)(n+1)$.

Answer 1

$S_1=\left(n/2\right)[2.1+(n-1\left)1\right]$,
$S_2=\left(n/2\right)[2.2+(n-1\left).2\right]$
$S_m=\left(n/2\right)[2.m+(n-1).m]$.
$\therefore S_1+S_2+...+S_3$
$=n(1+2+3+...m)+\frac{n(n-1)}{2}.(1+2+3+...m)$.
$=\frac{m(m+1)}{2}[n+\frac{n^2-n}{2}]$
$=\frac{m(m+1)}{2}\cdot \frac{n(n+1)}{2}=\frac{1}{4}mn(m+1)(n+1)$.
Here we have used
$1+2+3+...+m=\left(m/2\right)[1+m]$
i.e., $S=\left(n/2\right)[a+l]$.