Question

If $S_1,S_2,S_3,\dots S_m$ are the sums of $n$ terms of $mA.P.$ are $n$ and $m$
respectively, then find the sum of common differences are $1,3,5,\dots ,(2m-1)$ respectively, show that
$S_1+S_2+S_3+\dots S_m=\frac{1}{2}nm(nm+1)$

Answer 1

We have $S_1,S_2,S_3,.........S_m$ be the of n terms of an A.P's whose common difference are $1,3,5,.......(2p-1)$ respectively.

$S_1=\frac{n}{2}(2+(n-1\left)1\right)$

$S_2=\frac{n}{2}\left(2\right(2)+(n-1\left)3\right)$

$S_3=\frac{n}{2}\left(2\right(3)+(n-1\left)5\right)$

$S_m=\frac{n}{2}\left[2\right(m)+(n-1\left)\right(2m-1\left)\right]$

$\sum _1^m{S}_m=\frac{n}{2}\left[2\right(1+2+.....+m)+(n-1\left)\right(1+3+....+(2m-1)\left)\right]$

$=\frac{n}{2}[2\left(\frac{m.(m+1)}{2}\right)+(n-1\left)\frac{m}{2}\right(2+(m-1)2]$

$=\frac{mn}{2}(1+mn)$