If `S_1,S_2, S_3, S_m` are the sums of `n` terms of `m` A.P. `' s` whose first terms are `1,2,3,....m` and common differences are `1,3,5, ....(2m-1)` respectively. Show that `S_1+S_2.....+S_m=(m n)/2(m n+1)`

Explanation

`therefore 1,2,3,"..."p` are in AP.
Then,`2.1,2.2,2.3,"…",2p` are also in AP. `" " ".....(i)"`
`" " ["multiplying 2 to each term"]`
and `1,3,5,"…",(2p-1)` are in AP.
Then, `(n-1)*1,(n-1)*3,(n-1)*5,"....",(n-1)(2p-1)` are also in AP. `" " " ""....(ii)"`
`" " " " " [multiplying (n-1) to each term ]"`
From Eqs.(i) and (ii),we get
`2*1+(n-1)*1,2*2+(n-1)*3,2*3+(n-1)*5,"....",2p+(n-1)(2p-1)` are also in AP. `" " " " "....(iii)"`
`" " [" adding corresponding terms of Eqs.(i)and (ii)"]`
From Eq. (iii),
`(n)/(2){2*1+(n-1)*1},(n)/(2){2*2+(n-1)*3},(n)/(2){2*3+(n-1)*5},"...,"`
`(n)/(2){2p+(n-1)(2p-1)} " are also in AP"`
`" " " "["multiplying (n)/(2) to each term"]`
`i.e.S_(1),S_(2),S_(3),"...",S_(p)` are in AP.
`therefore S_(1)+S_(2)+S_(3)+"..."S_(p)=(p)/(2){S_(1)+S_(p)}`
`=(p)/(2){(n)/(2)[2*1+(n-1)*1]+(n)/(2)[2*p+(n-1)(2p-1)]}`
`=(np)/(2){2+(n-1)+2p+(n-1)(2p-1)}`
`=(np)/(4)(2np+2)=(1)/(2)np(np+1)`
Aliter
Here, `S_(1)=1+2+3+"..." "upto n terms" =(n(n+1))/(2),etc*`
`S_(2)=2+5+8+"..." "upto n terms" =(n)/(2)[2*2+(n-1)3]`
`=(n(3n+1))/(2)`
Similarly, `S_(3)=3+8+13+"..." "upto n terms" =(n(5n+1))/(2),etc`.
Now, `S_(1)+S_(2)+S_(3)+"..."S_(p)`
`=(n(n+1))/(2)+(n(3n+1))/(2)+(n(5n+1))/(2)+"..." "upto p terms"` lt brgt `=(n)/(2)[(n+3n +5n+"..." "upto p terms")+ (1+1+1+"...""upto p terms)"]`
`=(n)/(2)[(p)/(2)(2n+(p-1)2n)+p]`
`=(np)/(2)[n+n(p-1)+1]= (1)/(2)np(np+1)`