# Product of two terms inquiry

# 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 6*n* + 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:

Is it true or false?

If it is true, then is it true for some or all sequences?

3 x 5 = 15 and 3, 5 and 15 are in the sequence generated by 2

*n*+ 1.Does it only work with consecutive terms?

Are there any sequences when you could use the sum of the terms?

How could you show this is always true?

What would happen if you multiplied terms from two sequences? Would the answer be in both sequences?

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 2*n* + 2, 3*n* + 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 6*n* + 1** **is also a term in the same sequence. For example, the product of 7 and 13 (both terms in the sequence generated from 6*n* + 1) is 91, which is the fifteenth term in the sequence. The illustration below shows how George proved his conjecture.

## Counter-example

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

3*k* - 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] = 9k^{2 }+ 30k + 16. This can be written as 3(3*k*^{2} + 10*k*) + 16, which is not in the form 3*n* - 1.

## Proof

The prompt is true for any expression of the *n*^{th} term that has a constant of one (for example, 6*n* + 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 *a*^{2}. This can then be rearranged to give *a* as the final term. The product of the first two terms, for example, is *a*[(*ak*^{2} + 3*ak* + *a*) + *a*].