OxfordAQA International AS and A-level Mathematics: Binomial series for rational powers and series of rational functions (9660)

This study guide covers specification section P2.2 (Sequences and series) of the OxfordAQA International AS and A-level Mathematics (9660) specification, Version 5.2: the binomial series for any rational power, and the series expansion of rational functions using partial fractions. The specification states that this content is part of International A-level (A2) Unit P2 (Pure Maths) and is not examined at International AS. It applies to International A-level exams from May/June 2019.

Use it with the revision notes and the practice questions. Integer powers are taught in sequences, series and binomial expansion. The course hub is OxfordAQA A-level Mathematics and the printable checklist lists every outcome. If you are not sure where to start, take a free 10-minute diagnostic.

What this unit covers

Spec	What you must be able to do
P2.2	Expand (1 + x)ⁿ as a series for any rational n, with
P2.2	Expand (a + bx)ⁿ by taking out a factor, and state the range of x for which it is valid
P2.2	Expand a rational function as a series, including by using partial fractions

The specification gives the hardest binomial case as a power such as (2 + 3x)⁻², valid for |x| < 2/3. The hardest rational function it names has the shape (3 + 2x²) / ((2x + 1)(x − 3)²): a linear factor and a repeated linear factor. You do not need to handle irreducible quadratic factors in the denominator.

1. The binomial series for any rational n

For a positive integer n, (1 + x)ⁿ has n + 1 terms and is true for every x. When n is negative or a fraction, the expansion never stops. It is an infinite series, and it only equals (1 + x)ⁿ when |x| < 1.

(1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …     for |x| < 1

Here n! means n factorial: 2! = 2 and 3! = 6. The same pattern continues: each new term multiplies the previous coefficient by the next lower value of n and divides by the next integer.

Work with brackets. A fraction or a negative number must be bracketed every time it is multiplied, or the signs go wrong.

Worked example 1

Expand (1 − 3x)^(−1/2) in ascending powers of x up to and including x³. State the values of x for which the expansion is valid.

Write the expression as (1 + u)ⁿ with u = −3x and n = −1/2.

Term 1:  1
Term 2:  n u = (−1/2)(−3x) = 3x/2
Term 3:  n(n − 1)/2! · u² = (−1/2)(−3/2)/2 · 9x² = (3/8)(9x²) = 27x²/8
Term 4:  n(n − 1)(n − 2)/3! · u³ = (−1/2)(−3/2)(−5/2)/6 · (−27x³)
         = (−5/16)(−27x³) = 135x³/16

So (1 − 3x)^(−1/2) = 1 + 3x/2 + 27x²/8 + 135x³/16 + …

The series needs |u| < 1, so |−3x| < 1, which gives |x| < 1/3.

2. Expanding (a + bx)ⁿ

The series is for (1 + something)ⁿ only. If the first term is not 1, take it out:

(a + bx)ⁿ = aⁿ (1 + (b/a)x)ⁿ

Then expand with u = (b/a)x. The condition |u| < 1 becomes |x| < |a/b|. Remember to raise the constant a to the power n as well. With a fractional power, aⁿ is a root, so choose a only when it is simple to evaluate, or leave it as a surd.

Worked example 2

(a) Expand √(9 + 2x) up to the term in x³. (b) State the range of validity. (c) Use x = 0.1 to estimate √9.2.

√(9 + 2x) = (9 + 2x)^(1/2) = 9^(1/2) (1 + 2x/9)^(1/2) = 3 (1 + 2x/9)^(1/2)

Let u = 2x/9 and n = 1/2.

(1 + u)^(1/2) = 1 + u/2 − u²/8 + u³/16 − …

The third and fourth coefficients are (1/2)(−1/2)/2 = −1/8 and (1/2)(−1/2)(−3/2)/6 = 1/16. Substitute u:

= 1 + x/9 − (1/8)(4x²/81) + (1/16)(8x³/729)
= 1 + x/9 − x²/162 + x³/1458

Multiply by 3:

(a) √(9 + 2x) = 3 + x/3 − x²/54 + x³/486 + …

(b) |2x/9| < 1, so |x| < 9/2.

(c) x = 0.1 is inside the range. Then 9 + 2x = 9.2 and the series gives 3 + 0.03333… − 0.000185… + 0.0000002… ≈ 3.03315. Rounded to 5 decimal places, the series value is 3.03315 and √9.2 is 3.03315.

Estimates like this work best when x is small, because the higher powers shrink quickly.

Worked example 3

Expand (5 − 4x)⁻² up to the term in x² and state the range of validity.

(5 − 4x)⁻² = 5⁻² (1 − 4x/5)⁻² = (1/25)(1 + u)⁻²,   u = −4x/5
(1 + u)⁻² = 1 − 2u + 3u² − …
          = 1 + 8x/5 + 3(16x²/25)
          = 1 + 8x/5 + 48x²/25

Multiplying by 1/25: (5 − 4x)⁻² = 1/25 + 8x/125 + 48x²/625 + …

Validity: |4x/5| < 1, so |x| < 5/4.

The coefficients 1, −2, 3, −4 in (1 + u)⁻² are a quick check on any negative-integer power. In general (1 + u)⁻ᵏ starts 1 − ku + k(k + 1)/2 u².

Why the range matters

The series for (1 + x)⁻¹ is 1 − x + x² − x³ + …, a geometric series with first term 1 and common ratio −x. A geometric series has a sum only if the ratio has modulus less than 1, which is exactly |x| < 1. At x = 2 the terms are 1, −2, 4, −8, … and they grow, so the series cannot equal 1/3. The binomial series has the same limit for every rational n. Outside the range the series is not a valid representation of the function, whatever it looks like.

3. Products of a polynomial and a binomial series

When the expression is a product, expand the binomial part to one order higher than you think you need, then multiply out and collect like powers.

Worked example 4

Expand (3 + x)(1 + 4x)^(−1/2) up to the term in x³ and state the range of validity.

(1 + 4x)^(−1/2): u = 4x, n = −1/2
= 1 + (−1/2)(4x) + (−1/2)(−3/2)/2 (16x²) + (−1/2)(−3/2)(−5/2)/6 (64x³)
= 1 − 2x + 6x² − 20x³

Multiply by (3 + x):

3(1 − 2x + 6x² − 20x³) = 3 − 6x + 18x² − 60x³
x(1 − 2x + 6x² − …)   =     x −  2x² +  6x³
Sum:                     3 − 5x + 16x² − 54x³

The expansion is 3 − 5x + 16x² − 54x³ + …. The factor (3 + x) is a polynomial and has no restriction, so the range comes from 1 + 4x only: |4x| < 1, so |x| < 1/4.

4. Series of rational functions

A rational function is a fraction of two polynomials. If the denominator is a product of factors, a single binomial series cannot be used. Split the fraction into partial fractions, expand each part, and add.

Method:

1. Check the numerator has lower degree than the denominator (the cases in the specification do).
2. Write the partial-fraction form. A repeated factor (2 − x)² needs both B/(2 − x) and C/(2 − x)².
3. Find the constants by substitution and comparing coefficients.
4. Rewrite each part as a constant times (1 + u)ⁿ and expand.
5. Add the series term by term.
6. The result is valid only where every part is valid. Take the narrowest range.

Worked example 5

Express f(x) = (13 + 6x − x²) / ((1 + 3x)(2 − x)²) in partial fractions. Hence expand f(x) up to the term in x³ and state the range of validity.

f(x) = A/(1 + 3x) + B/(2 − x) + C/(2 − x)²
13 + 6x − x² = A(2 − x)² + B(1 + 3x)(2 − x) + C(1 + 3x)

Put x = −1/3: 13 − 2 − 1/9 = A(7/3)², so 98/9 = 49A/9, which gives A = 2. Put x = 2: 13 + 12 − 4 = 21 = 7C, so C = 3. Compare x²: −1 = A − 3B, so −1 = 2 − 3B and B = 1.

So f(x) = 2/(1 + 3x) + 1/(2 − x) + 3/(2 − x)².

Expand each part:

2/(1 + 3x)     = 2(1 − 3x + 9x² − 27x³)           = 2 − 6x + 18x² − 54x³
1/(2 − x)      = (1/2)(1 − x/2)⁻¹ = (1/2)(1 + x/2 + x²/4 + x³/8)
               = 1/2 + x/4 + x²/8 + x³/16
3/(2 − x)²     = (3/4)(1 − x/2)⁻² = (3/4)(1 + x + 3x²/4 + x³/2)
               = 3/4 + 3x/4 + 9x²/16 + 3x³/8

Here (1 − x/2)⁻² uses 1 + 2u + 3u² + 4u³ with u = x/2.

Add the columns:

constant: 2 + 1/2 + 3/4 = 13/4
x:        −6 + 1/4 + 3/4 = −5
x²:       18 + 1/8 + 9/16 = 299/16
x³:       −54 + 1/16 + 3/8 = −857/16

f(x) = 13/4 − 5x + 299x²/16 − 857x³/16 + …

Validity: the first part needs |3x| < 1, so |x| < 1/3. The others need |x/2| < 1, so |x| < 2. Both must hold, so |x| < 1/3.

A quick check: f(0) = 13/(1 × 4) = 13/4, which matches the constant term.

Factors written as (x − 3)

A denominator such as (x − 3)² is the same as (3 − x)², so 1/(x − 3)² = 1/(9(1 − x/3)²) = (1/9)(1 + 2x/3 + x²/3 + …). A single factor flips its sign: 1/(x − 3) = −1/(3 − x). Always put the constant first and the x-term second, so each part can be written as a constant times (1 + u)ⁿ.

For the specification’s own example, (3 + 2x²)/((2x + 1)(x − 3)²) = 2/(7(1 + 2x)) − 6/(7(3 − x)) + 3/(3 − x)². Expanding each part and adding gives 1/3 − 4x/9 + 11x²/9 + …, valid for |x| < 1/2. The specification’s binomial example is (2 + 3x)⁻² = (1/4)(1 + 3x/2)⁻² = 1/4 − 3x/4 + 27x²/16 − …, valid for |x| < 2/3.

Common errors

• Forgetting to raise the constant to the power. In (4 + x)^(−1/2) the leading factor is 4^(−1/2) = 1/2, not 4.
• Dropping a bracket on a negative u. In (1 − 3x)^(−1/2) the u² term uses (−3x)² = +9x², and the u³ term is negative.
• Giving the validity as |x| < 1 for every expression. It is |u| < 1, so for (1 + 4x) it is |x| < 1/4.
• Stopping at the first partial fraction found, or using only B/(2 − x) for a squared factor.
• Mixing up the sign of the series for a factor written as (x − 3).
• Stating the range of one partial fraction rather than the narrowest range for all of them.

Next steps

Try the practice questions, check the key results in the revision notes, and tick the outcome off on the checklist. Take the diagnostic to find your weakest area.

Official syllabus

OxfordAQA International AS and A-level Mathematics (9660), Version 5.2, section P2.2 (International AS exams from May/June 2018, A-level from May/June 2019), published by OxfordAQA.