The series of natural numbers is divided into groups`:(1);(2,3,4);(5,6,7,8,9)...` and so on. The sum of numbers in the `n^(th)` group is

Explanation

The groups are (1), (2, 3, 4), 95, 6, 7, 8, 9)...
The number of terms in the groups are 1, 3, 5....
`:.` The number of terms in the `n^(th)` group `= (2n -1)` the last terms of the `n^(th)` group is `n^(2)`
If we count from last term common difference should be `-1`
So the sum of numbers in the `n^(th)` group `= ((2n - 1)/(2)) {2n^(2) + (2n - 2) (-1)}`
`= (2n - 1) (n^(2) - n + 1) = 2n^(3) - 3n^(2) + 3n - 1= n^(3) + (n -1)^(3)`