## New at GCSE: Quadratic sequences

An item of content newly introduced into GCSE (though it has come and gone over the years) is finding the *n*th terms of quadratic sequences. Expressing general rules that relate the *n*th terms of pattern sequences to their values is a way to harness pupils’ natural pattern-seeking to develop their understanding of, and fluency with, algebra. (See: Key Ideas in Teaching Mathematics, Anne Watson, Keith Jones, Dave Pratt, 2013, Oxford University Press, Page 31).

Here are some examples of tasks that provide opportunities for pupils to see visual patterns in various different ways, and thereby arrive at equivalent quadratic expressions for the *n*th term. Alternatively pupils could count the number of squares in each image to obtain a numerical sequence, and then use an algebraic method to derive an expression for the *n*th term. Usually the numerical-sequence method is more laborious and is often used without pupils understanding why it works. Therefore pupils should appreciate that, with sequences of images it is often easiest to use the geometric structure of the images - to see how the images ‘grow’ to form the sequence - so that when they arrive at an expression for the *n*th term they know precisely why it must be correct.

In all the following examples the task is to

*find an expression for the number of squares in the nth image*

**Example 1**

Pupils may see each image as a square with side-length *n* together with a rectangle of height 1, and length *n*:

The number of squares in the *n*th image = *n*^{2} + *n*.

A different way to see each image is as a rectangle with side-lengths *n* and 2*n* with a smaller rectangle of side-lengths *n* and *n* – 1 removed from it:

The number of squares in the *n*th image |
= *n* × 2*n* – *n*(*n* – 1)
= 2*n*^{2} – *n*^{2} + *n*
= *n*^{2} + *n* |

The numerical approach entails counting the number of squares in each of the first few images and then working with the numerical sequence so produced. Pupils first have to understand that the general term of a quadratic sequence is of the form a*n*^{2} + b*n* + c. They can then use reasoning to deduce the values of a, b and c for the numerical sequence they have formed, as follows.

First write the sequence of number of squares in each image, and underneath write the first differences, and below that write the second differences. Then write the same first few terms algebraically by substituting 1, then 2, then 3, and so on, in turn for *n* in a*n*^{2} + b*n* + c. Also write the first and second differences algebraically.

By comparing the second differences expressed algebraically (2a) with the numerical second differences pupils will see that, for this sequence, 2a = 2, so a = 1.

By equating any of the first differences expressed algebraically with the corresponding numerical first difference pupils can now find the value of b. For example, 3a + b = 4. Since a = 1, 3 + b = 4, so b = 1.

To find the value of c pupils have to understand that when *n* = 0, the term expressed algebraically is just c, since a*n*^{2} and b*n* are both 0. By extending the sequence of first differences backwards (in this example to get 2, shown in brackets above) they can extend the quadratic sequence itself back to arrive (in this example) at 0 as the term when *n* = 0.

Therefore the *n*th term, a*n*^{2} + b*n* + c, is *n*^{2} + *n*.

Pupils can compare the relatively complex third method to the previous direct ways of seeing the *n*th term using the structure of the images.

It is helpful to remind pupils that usually the images can be cut up into a combination of rectangles and squares, or every image can be seen as a ‘surrounding’ rectangle from which one or more rectangles have been removed. They then just have to decide how the side-lengths of the squares and rectangles composing, or removed from, an image are related to the position of the image in the sequence. Here is another example in which the images can be cut up into squares and rectangles in various ways.

**Example 2**

Pupils could imagine cutting each image into a rectangle of side-lengths *n* + 2 and 2*n* and a unit square, as shown below:

The number of squares in the *n*th image |
= 2*n*(*n* + 2) + 1
= 2*n*^{2} + 4*n* + 1 |

Or pupils may see a rectangle with side-lengths 2*n* + 1 and *n* + 2 from which a rectangle of unit width and height *n* + 1 has been removed, as shown below:

The number of squares in the *n*th image |
= (2*n* + 1)(*n* + 2) – (*n* + 1)
= 2*n*^{2} + 5*n* + 2 – *n* – 1
= 2*n*^{2} + 4*n* + 1 |

Or they may see each image as two *n*-by-*n* squares of area *n*^{2} plus four *n*-by-1 rectangles of area *n* plus a unit square, possibly as shown below:

The number of squares in the *n*th image |
= *n*^{2} + *n*^{2} + *n *+* n* + *n* + *n* + 1
= 2*n*^{2} + 4*n* + 1 |

Pupils can now compare the relatively simple way in which they found the expression for the *n*th term by seeing a structure in the images with the more laborious procedure of counting the squares in the first few images and working algebraically on the resulting numerical sequence, as shown below:

By equating the second differences 2a |
= 4 |

a |
= 2 |

By equating the first differences 3a + b |
= 10 |

Since a = 2, 6 + b |
= 10 |

b |
= 4 |

By extending the sequence back to *n* = 0, using the pattern of first differences,

when *n* = 0, a*n*^{2} + b*n* + c = c = 1

Therefore the *n*th term a*n*^{2} + b*n* + c = 2*n*^{2} + 4*n* + 1.

Pupils may enjoy devising their own examples for others to solve. In the meantime here are some more.

Once you have a jutification for the sequence 'continuing in the same way', Colin's description of different ways of treating George Boole's method of constant k'th differences is really really helpful! It could save a lot of students a lot of bother if their techer has considered these different ways of exploiting constant k'th differences!