A) The major flaw in your argument is that you have used the "first term" in two different ways, initially as your zeroth term, and then "by letting n = 1 and using the first term". So the answer to your question is yes, your students would be disadvantaged by this method.

You may have made a mistake in your question and actually meant to assign the second term to solve the problem. What you suggest is confusing, especially with a middle set, and this has come across in the question.

Students in this ability group need to know that what is being suggested makes common sense - Jcalling the first term the zeroth term might not make sense to them.

The general formula for a quadratic sequence is Un = an2 + bn + c. In the Framework, the notation suggested is Tn=an2+bn+c. After a discussion with Teresa Smart, from the University of London Institute of Education, about which notation to use, we concluded that, as long as the notation is defined, using "U" or "T" makes no difference.

Here is a method that your students might find helpful. I shall focus on the sequence: 1, 7, 17, 31, 49 I For those who do not know how to create the general formula for a sequence, I will take you through the steps: First, find the difference between consecutive members of the sequence: 17 - 7 gives a difference of 10; 31 - 17 gives a difference of 14, and so on.

These make the "first set of differences". If these differences had all been the same value, I would have known that I had a linear sequence (graphically a straight line) for the nth term: Un = bn + c. But in this case, the differences are not the same.

When the "first set of differences" are different, we calculate the "second set of differences" by looking at the consecutive members of the first set (6, 10, 14, 18 I). In this case, 10 - 6 = 4, 14 - 10 = 4, and so on; all the second differences are the same and equal to 4. The fact that the "second set of differences" all have the same value tells us that this is a quadratic sequence (graphically a parabola): Un = an2 + bn + c.

In a quadratic sequence the matching value found in the second set of differences is always twice the coefficient "a" in front of the squared term in the sequence. Therefore by halving this second difference we can find the value of "a" in this case, a = 2 (4 V 2).

We can immediately rewrite our nth term by replacing the "a" with the value 2: Un = 2n2 + bn + c.

= To find the value for "c" we can take the sequence back to the term before the first term, which is U0.

As you can see in the diagram, U0 is - 1. So we can substitute into the general formula U0 = 2 X 0 + b X 0 + C.

U0 = C, but we already know from carrying the sequence backwards that this term is equal to - 1, thus C = - 1, which leads to Un = 2n2 + bn - 1.

Next, the value of the first term is 1 (U1 = 1), so we use this to find a value for "b":

1 = 2 X 12 + b X 1 - 1

1 = 2 + b - 1

0 = b

The general term of this quadratic sequence is therefore Un = 2n2 - 1.

Students should check that their solution is correct by trying to work out a term - perhaps the fourth term - to see if it matches their sequence: U4 = 2 X 42 - 1 = 32 - 1 = 31

* This is, indeed, the fourth term of the sequence.

Wendy Fortescue-Hubbard is a teacher and game inventor. She has been awarded a three-year fellowship by the National Endowment for Science, Technology and the Arts (NESTA) to spread maths to the masses.

www.nesta.org.uk

Email your questions to Mathagony Aunt at teacher@tes.co.uk Or write to TES Teacher, Admiral House, 66-68 East Smithfield, London E1W 1BX