Sums of Successive Integers

Suppose we want to find 1 + 2 + ... + 100.

Here's a neat trick, attributed to Gauss when he was a child. His teacher told his class they could go to outside and play after they'd summed the numbers between 1 and 100. A few moments later, young Gauss got up to leave. The teacher asked where he was going. Gauss said out to play. What's the sum, asked the teacher. 5050 Gauss replied. Ummm, said the teacher. Here's whay Gauss had done.

1 + 100 = 101 2 + 99 = 101 3 + 98 = 101 4 + 97 = 101 ... ... ... 99 + 2 = 101 100 + 1 = 101

The two left columns sum to 2⋅(1 + 2 + ... + 100), and the right column sums to 100⋅101, so

2⋅(1 + 2 + ... + 100) = 100⋅101

That is, 1 + 2 + ... + 100 = 100⋅101/2

In general, 1 + 2 + ... + N = N⋅(N+1)/2

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