# The prompt

Mathematical inquiry processes: Explore; generate examples and counter-examples; generalise and prove. Conceptual field of inquiry: Position-to-term rules; algebraic expressions.

The prompt was devised by Helen Hindle, head of the mathematics department at Park View School (Haringey, UK). It generalises from a student's observation made during the intersecting sequences inquiry. The student proved the statement is true for the sequence generated from 6n + 1. Teachers could use the prompt as the start of a stand-alone inquiry or introduce it as a separate pathway during the intersecting sequences inquiry.

The questions and observations that develop from the statement include:

Students realise that most expressions do not generate arithmetic sequences for which the prompt is true. During exploration, they conclude that the prompt is true for any expression with a constant of one - that is, ending with +1. It is also true for any expression in the form an + a, such as 2n + 2, 3n + 3 and so on.

# Conjecture, counter-example and proof

## Conjecture and proof

George Marsden (a year 10 student at St. Andrew's School, Leatherhead, UK) proved his conjecture that the product of any two terms in the sequence given by the expression for the nth term 6n + 1 is also a term in the same sequence. For example, the product of 7 and 13 (both terms in the sequence generated from 6n + 1) is 91, which is the fifteenth term in the sequence. The illustration below shows how George proved his conjecture.

## Counter-example

Using 3n - 1 as the expression for the nth term, we get 2, 5, 8, 11, 14. In a general form, we have:

3k - 1, 3(k + 1) - 1, 3(k + 2) - 1, 3(k + 3) - 1.

The product of, say, the second and fourth term is [3(k + 1) - 1][3(k + 3) - 1] = 9k2 + 30k + 16. This can be written as 3(3k2 + 10k) + 16, which is not in the form 3n - 1.

## Proof

The prompt is true for any expression of the nth term that has a constant of one (for example, 6n + 1). The general sequence starts:

ak + 1, a(k + 1) + 1, a(k + 2) + 1, a(k + 3) + 1.

Whichever two terms we choose, the expression for the product of the two will always end with +1, which corresponds to the general form of an + 1.

The prompt is also true when the coefficient and constant in the expression are the same. The general sequence starts:

ak + a, a(k + 1) + a, a(k + 2) + a, a(k + 3) + a.

Whichever two terms we choose, the expression for the product of the two will always end with a term in a2. This can then be rearranged to give a as the final term. The product of the first two terms, for example, is a[(ak2 + 3ak + a) + a].