Consider $$a_{1}=1, a_{2}=2,a_{3}=4,a_{4}=7,a_{5}=11.......a_{n-1}, a_{n}$$
Now, $$a_{2} - a_{1} = 1 ....(1)$$
$$a_{3} - a_{2} = 2.......(2)$$
$$a_{4} - a_{3} = 3 ......(3)$$
$$a_{5} - a_{4} = 4.......(4)$$
$$......a_{n-1} - a_{n-2} = n-2.......(n-2)$$
$$a_{n} - a_{n-1} = n-1 ............(n-1)$$
Adding left and right side of all equations we get, $$a_{n} - a_{1} = (n-1)(n)/2$$
$$a_{n} = (n-1)(n)/2 + a_{1} = \frac{\left(n^2-n+2\right)}{2}$$
Sum of the series = (Sum of ($$n^2$$) - Sum of (n) +Sum of (2))*(1/2)
=$$\left(\frac{n\left(n+1\right)\left(2n+1\right)}{6}-\frac{n\left(n+1\right)}{2}+2n\right)\cdot\left(\frac{1}{2}\right)$$
= $$\frac{1}{2} \times (Sum of squares of first 30 natural numbers) \times (Sum of first 30 natural numbers)$$
So, sum of the first 30 terms is 30*31*61/12 - 30*31/4 + 30 = 4525
