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
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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