Sequences and Series
In AS/A-Level Mathematics, a sequence is a list of ordered numbers found by applying a formula to the nth term.
Recurrence Relationships
In AS/A-Level Mathematics, this is where the next term of a sequence is defined using the previous term(s). To define a recurrence relation, you have to give the first term. Then you give a formula to tell you how to work out the next term from the previous ones.
For example, consider the sequence: 2, 4, 8, 16, 32, ... . Each term in the sequence is got by doubling the previous term. So to define the recurrence relation, we give the first term, written U1 = 2. Then we write: Un = 2(Un-1). This just means that the nth term, Un is equal to 2 × the (n-1)th term, Un-1. An example of a recurrence relationship is Un = Un-1 + 5.
Series
The series of a sequence is the sum of the sequence to a certain number of terms. It is often written as Sn. So if the sequence is 2, 4, 6, 8, 10, ... , the sum to 3 terms = S3 = 2 + 4 + 6 = 12.
The Sigma Notation
The Greek capital sigma, written S, is usually used to represent the sum of a sequence. This is best explained using an example:
This means replace the r in the expression by 1 and write down what you get. Then replace r by 2 and write down what you get. Keep doing this until you get to 4, since this is the number above the Sigma. Now add up all of the term that you have written down.
This sum is therefore equal to (3×1) + (3×2) + (3×3) + (3×4) = 3 + 6 + 9 + 12 = 30.
The General Case
This is the general case. For the sequence U, this means the sum of the terms obtained by substituting in 1, 2, 3,... up to and including n in turn for r in Ur. In the above example, Ur = 3r + 2 and n = 3.
Arithmetic Progressions
An arithmetic progression is a sequence where each term is a certain number larger than the previous term. The terms in the sequence are said to increase by a common difference, d. For example: 3, 5, 7, 9, 11, is an arithmetic progression where d = 2. The nth term of this sequence is 2n + 1 .
In general, the nth term of an arithmetic progression, with first term a and common difference d, is:
The Sum to n terms of an Arithmetic Progression
This is given by:
This formula is derived as follows:
The sum to n terms is given by: Sn = a + (a + d) + (a + 2d) + … + (a + (n – 1)d) ①
Written backwards, we get: Sn = (a + (n – 1)d) + (a + (n – 2)d) + … + a ②
Now let’s add ① and ②
2Sn = [2a + (n – 1)d] + [2a + (n – 1)d] + … + [2a + (n – 1)d]
So Sn = ½ n [2a + (n – 1)d]
That's all!
