Sum of the first n terms:
- Sum of the first n natural numbers is $$\frac{n(n+1)}{2}$$
- Sum of the squares of the first n natural numbers is $$\frac{n(n+1)(2n+1)}{6}$$
- The sum of cubes of first 'n' natural numbers = $$(\frac{n(n+1)}{2})^{2}$$
- The sum of first 'n' odd natural numbers= $$n^{2}$$
- The sum of first 'n' even natural numbers= n(n+1)
- In any series, if the sum of first n terms is given by $$S_{n}$$,then the $$n^{th}$$ term $$T_{n}=S_{n}-S_{n-1}$$
