Question

please prove this question from AP

Answer 1

For Sk we say the first term is k and the common difference is 2k-1. So its sum upto n terms shall be n/2{2k + (n-1)(2k-1)} = nk + n(n-1)(2k-1)/2. Now, S1 + S2 + S3 + ........+Sm = n(1+2+3+4+5+6+......+m) + n(n-1){1+3+5+.....+(2m-1)}/2. 

Now 1,2,3.....m is an A.P. with first term 1 and last term m and total number of terms m. So, 1+2+3+4+....+m = m/2(1+m)

Similarly,1+3+5+7+.......+(2m-1) = m^2.

So our LHS now equals nm(m+1)/2 +n(n-1)m^2/2
                                     =1/2{mn(m+1) + n(n-1)m^2}
                                     =1/2(nm^2 + mn m^2n^2 - nm^2)
                                     =1/2(mn+m^2n^2)
                                     =mn(mn+1)/2 as desired.