EMT 725

Philippa M. Rhodes

Derive and prove a formula for the sum of the first n integers.

Derivation I

Make a triangular array. This one represents the sum of the first 6 integers.

Duplicate the array and place the copy so that a rectangular shape is formed.

In this example, we have 2 * (the sum of the first 6 integers) = 6 * 7. After other examples, we can conclude that 2 * (the sum of the first n integers) = the area of the rectangle = n (n + 1). So,

= .

Derivation II

We want to know 1 + 2 + 3 + . . . + (n - 2) + (n - 1) + n.

If n is even, then we can pair 1 with n, 2 with (n-1), 3 with (n-2), etc. The sums of each of these pairs is n + 1. Since n is even, we have n/2 such pairs. Therefore,

1 + 2 + 3 + . . . + (n - 2) + (n - 1) + n = .

If n is odd, then we can pair 1 with (n - 1), 2 with (n - 2), etc. This will leave n by itself. The sums of each of the (n-1)/2 pairs will equal n. Therefore,

1 + 2 + 3 + . . . + (n - 2) + (n - 1) + n = = =

Again, we have

=

Proof by induction

Let n = 1. So, 1 = 1 * 2 / 2. Thus, the statement holds when n = 1.

Assume the statement is true for n = k. So, 1 + 2 + 3 + . . . + k = k * (k+1) / 2.

1 + 2 + . . . + k + (k + 1) = k * (k+1) / 2 + (k +1) =

=.
So, the statement is true for n = k +1.
By the Principle of Mathematical Induction, the statement holds for any natural number, n.