Prove by induction that the sum of the first n integers can be written as (1/2)(n)(n+1).

For n = 1, the sum is given by (1/2)(1)(1+1), which gives 1, the expected result. We now assume that the statement is true for some k. If we look at k+1, the sum is given by 1 + 2 + ... + k + (k+1). Since we have assumed that 1 + 2 + ... + k = (1/2)(k)(k+1), this can be rewritten as (1/2)(k)(k+1) + (k+1). Simplifying this gives (1/2)(k+1)(k+2), which is the required result. If the statement is true for n = k, we have shown it to be true for n = k + 1. Since the statement is true for n = 1, it is shown to be true for all n >= 1.