Let Tn be the nth term and Sn be the sum to n terms of the given series.
We have,
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn … (1)
Equation (1) can be rewritten as:
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn ……..(2)
By subtracting (2) from (1) we get
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn
Sn = 1 + 3 + 6 + 10 + 15 + …………. + Tn-1 + Tn
0 = 1 + [2 + 3 + 4 + 5 + … + (Tn – Tn-1)] – Tn
The difference between the successive terms are 2, 3, 4, 5
So these differences are in A.P
Now,

∴ The sum of the series is n/6 (n + 1) (n + 2)