Sum of the First n Positive Integers: Another Solution

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

sum-of-integers

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

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

sum-of-1-up-to-n

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

2S = n(n+1)

which implies that

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

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