How about the sequence of integers from 1 through 99. 1, 2, 3, ... , 97, 98, 99 To sum this sequence, let that sum be S, S = 1 + 2 + 3 + ... + 97 + 98 + 99 Then write the same sum in reverse order underneath it, and add the two equations: S = 1 + 2 + 3 + ... + 97 + 98 + 99 S = 99 + 98 97 + ... + 3 + 2 + 1 -------------------------------------------- 2S = 100 + 100 + 100 + ... + 100 + 100 + 100 2S = 99*100 2S = 9900 S = 4950 So the sum of the integers from 1 through 99 is 4950. Edwin