Late last month, we have seen how Gauss had possibly calculated the sum of the first *n* positive integers. In this post, we explore another solution. As we have done before, let us first start with a specific example, and then generalize later. What is the sum of the first 6 positive integers?

If we let S be the sum of the first 6 positive integers, then S = 1 + 2 + 3 + 4 + 5 + 6. Since the order of the addends does not change the sum, S = 6 + 5 + 4 + 3 + 2 + 1. Adding the two equation, we have

so, 2*S* = 6(7) = 42 and *S* = 21.

We can generalize the addition above as shown in the following image.

As we can see, each sum is equal to *n* + 1 and there are *n* of them. Therefore,

$latex 2S = n(n+1)$

which implies that

$latex S = \displaystyle\frac{n(n+1)}{2}$

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