The odd positive integers are grouped as shown below:-

{1},{3,5}{7,9,11}(13,15,17,19}...

Find the sum of the numbers in the n^th set...

n³

Closely viewing we get

Sum of first set = 1³

Sum of second set = 2³

Sum of third set = 3³

hence it continues till n^th set hence

we get sum as n³

yeah even i got it by observation and by a method but i am not sure about one of the steps so was looking for  a method..

If you want it really rigorous, here goes:

The one thing I want to assume is that the number of terms in the n-th bracket is n.

I claim that the first term in each bracket is n²-n+1. You can prove it by looking at the series 1, 3, 7 and using finite differences or use induction. If the first term in a the kth bracket is k²-k+1, since there are k consecutive odd numbers the last term will be k² -k+2k-1. Hence the (k+1)th bracket will begin with k²-k+2k+1 = k²+k+1 = (k+1)²-(k+1)+1 and so our claim is correct

So we have to sum the terms (n²-n+1)+(n²-n+3)+....+(n²-n+2n-1) = n(n²-n) + (1+3+5+...+2n-1)

= n³-n²+n² = n³