## Discrete Mathematics with Applications 4th Edition

The sum of the first n integers is given by: 1 + 2 + ··· + n = $\frac{n(n+1)}{2}$ 3 + 4 + 5 + 6 + ··· + 1000 = (1 + 2 + 3 + 4 + ··· +1000) − (1 + 2) = $\frac{1000\times(1001)}{2}$ -3 =500500 - 3 = 500,497