The sum of the first n integers starting with 1 is equal to `S_n = (n*(n+1))/2` . This can be proved by induction.

First, verify the formula for n = 1. `S_1 = 1` and `(1*2)/2` is also equal to 1.

Now prove that the relation is true for n + 1 if it is assumed to be true for n.

`S_(n+1) = S_n + n+1`

It has been assumed that `S_n = (n*(n+1))/2`

`S_(n+1) = S_n + n+1 `

=> `(n*(n+1))/2 + n + 1`

=> `(n*(n+1))/2 + (2*(n + 1))/2`

=> `(n+1)/2*(n+2)`

=> `((n+1)(n+2))/2`

As the formula holds for n = 1 and if it is assumed to be true for any number n it is also true for n + 1, the formula holds for all integral values of n.

The sum of the first n integers is given by the formula `S_n = (n*(n+1))/2`