# Every Third Triangular Number

This is a quick proof based on an observation inspired by “Mathematical Lens” in the May 2016 Mathematics Teacher (“Fence Posts and Rails” by Roger Turton). A triangular number is the sum of all integers from 1 to n. The general formula for T(n), the nth triangular number, is $T(n) = \frac{(n)(n + 1)}{2}$

Challenge: Prove that all triangular numbers are either multiples of three, or one more than a multiple of three. (That is, there are no triangular numbers that are one less than a multiple of three.)

We have three possible cases: $$n$$ is a multiple of 3, $$n + 1$$ is a multiple of 3, or $$n + 2$$ is a multiple of 3. Let $$k$$ represent some other integer. If the first case, $$n = 3k$$, so $$(n)(n + 1) = 3k(3k + 1)$$, which is obviously a multiple of 3. In the second case, $$n + 1 = 3k$$, so $$(n)(n + 1) = 3k(3k – 1)$$, which is obviously a multiple of 3.

This leaves the third case, in which neither $$n$$ nor $$n + 1$$ is a multiple of 3. In this case, since $$n + 2 = 3k$$, then $$n = 3k – 2$$ and $$n + 1 = 3k – 1$$. This means that $$2T(3k-2) = (3k – 1)(3k – 2) = 9k^2 – 9k + 2 = 9k(k – 1) + 2$$. Hence, $$T(3k-2) = 9\frac{k(k – 1)}{2} + 1$$. The first term here is a multiple of 3. Thus: $T(3k) = 3\frac{k(3k + 1)}{2} \\ T(3k-1) = 3\frac{k(3k – 1)}{2} \\ T(3k-2) = 3\frac{3k(k-1)}{2} + 1$

Q.E.D.

Clio Corvid

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