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.
=.
So, the statement is true for n = k +1.
By the Principle of Mathematical Induction, the statement holds for any
natural number, n. Return to Philippa's EMT 725 Page
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