The binomial series for Edexcel IGCSE Further Pure Mathematics: Methods and Worked Examples

The binomial series extends the familiar binomial expansion into territory where the exponent is no longer a positive integer, and that extension is where the real exam challenge lies

You may already know the binomial expansion for (1 + x)ⁿ when n is a positive integer. That expansion terminates after n + 1 terms and works for all values of x. What the edexcel igcse further pure mathematics specification adds is the expansion when n is rational (a fraction or a negative number). In that case the expansion is infinite, and it is valid only when |x| < 1. Understanding that distinction, and being able to apply the expansion confidently in both cases, is essential for the exam.

These edexcel igcse further pure mathematics the binomial series revision notes cover the entire section with clear worked examples, validity conditions, and the most common mistakes students make. As an edexcel igcse further pure mathematics explained resource, the approach is clean and methodical: learn the formula, see it applied, then practise until the pattern becomes automatic.

The binomial expansion for positive integer n

When n is a positive integer, the binomial expansion of (1 + x)ⁿ is:

(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + ... + xⁿ

This expansion has exactly n + 1 terms and is valid for all values of x.

For (a + b)ⁿ, you can factor out aⁿ and write it as aⁿ(1 + b/a)ⁿ, then expand.

Worked example: expanding (1 + x)⁵

(1 + x)⁵ = 1 + 5x + 5(4)x²/2! + 5(4)(3)x³/3! + 5(4)(3)(2)x⁴/4! + x⁵

= 1 + 5x + 10x² + 10x³ + 5x⁴ + x⁵.

Worked example: finding a specific term

Find the coefficient of x³ in the expansion of (2 + 3x)⁶.

The general term in (a + b)ⁿ is C(n, r) a^n-r b^r.

Here a = 2, b = 3x, n = 6, r = 3.

Term = C(6, 3) (2)³ (3x)³ = 20 x 8 x 27x³ = 4320x³.

The coefficient of x³ is 4320.

The binomial expansion for rational n

When n is not a positive integer (for example, n = 1/2, n = -1, n = -3/2), the expansion of (1 + x)ⁿ becomes an infinite series:

(1 + x)ⁿ = 1 + nx + n(n-1)x²/2! + n(n-1)(n-2)x³/3! + ...

This series is valid only when |x| < 1.

Why the validity condition matters

When n is a positive integer, the expansion terminates, so there is no convergence issue. When n is rational, the expansion has infinitely many terms, and the partial sums only converge to the correct value if |x| < 1. The exam tests this directly: you may be asked to state the range of values of x for which the expansion is valid.

Worked example: expanding (1 + x)^-2

Expand (1 + x)^-2 up to and including the term in x³, stating the values of x for which the expansion is valid.

Using the formula with n = -2:

1 + (-2)x + (-2)(-3)x²/2! + (-2)(-3)(-4)x³/3!

= 1 - 2x + 3x² - 4x³ + ...

Valid for |x| < 1.

Worked example: expanding (1 + x)^1/2

Expand (1 + x)^1/2 up to and including the term in x³.

n = 1/2:

1 + (1/2)x + (1/2)(-1/2)x²/2! + (1/2)(-1/2)(-3/2)x³/3!

= 1 + x/2 + (-1/4)x²/2 + (3/8)x³/6

= 1 + x/2 - x²/8 + x³/16.

Valid for |x| < 1.

Worked example: expanding (4 + x)^1/2

Expand (4 + x)^1/2 up to the term in x², and state the values of x for which the expansion is valid.

Factor out 4^1/2 = 2: (4 + x)^1/2 = 2(1 + x/4)^1/2.

Now expand (1 + x/4)^1/2 using n = 1/2 and replacing x with x/4:

= 1 + (1/2)(x/4) + (1/2)(-1/2)(x/4)²/2! + ...

= 1 + x/8 - x²/128 + ...

Multiply by 2: (4 + x)^1/2 = 2 + x/4 - x²/64 + ...

Valid when |x/4| < 1, i.e., |x| < 4.

Worked example: using binomial expansion to approximate a value

Use the expansion of (1 + x)^1/2 up to the x² term to estimate \u221a1.08.

Set x = 0.08: (1 + 0.08)^1/2 = 1 + (0.08)/2 - (0.08)²/8 = 1 + 0.04 - 0.0008 = 1.0392.

Calculator value: \u221a1.08 = 1.03923..., so the approximation is accurate to 4 decimal places.

Combining the binomial expansion with partial fractions

Some exam questions give you a rational expression such as (5x + 1) / ((1 + x)(1 - 2x)) and ask you to expand it as a series. The approach is to split into partial fractions first, then expand each fraction separately using the binomial series.

Worked example: Express 1 / ((1 + x)(1 - x)) in partial fractions and expand up to the term in x².

Partial fractions: 1 / ((1 + x)(1 - x)) = A/(1 + x) + B/(1 - x). Covering up: A = 1/2, B = 1/2.

So 1 / ((1 + x)(1 - x)) = (1/2)(1 + x)^-1 + (1/2)(1 - x)^-1.

(1 + x)^-1 = 1 - x + x² - ... (valid for |x| < 1).

(1 - x)^-1 = 1 + x + x² + ... (valid for |x| < 1).

Adding: (1/2)(1 - x + x²) + (1/2)(1 + x + x²) = 1 + x².

This makes sense because 1/((1 + x)(1 - x)) = 1/(1 - x²) = (1 - x²)^-1 = 1 + x² + x⁴ + ...

Practice questions

These edexcel igcse further pure mathematics practice questions on the binomial series will test your fluency. Work through each one before checking the solutions.

1. Expand (1 - 2x)^-3 up to and including the term in x³. State the values of x for which the expansion is valid.
2. Find the first four terms of the expansion of (8 + 3x)^1/3 in ascending powers of x.
3. Find the coefficient of x⁴ in the expansion of (1 + x)⁸.
4. Use the binomial expansion to find an approximation for (0.98)^1/2 correct to 5 decimal places.
5. Expand (1 + x)^-1(1 + 2x)^-1 up to the term in x² by first splitting into partial fractions.

Solutions

Question 1: (1 - 2x)^-3 = 1 + (-3)(-2x) + (-3)(-4)(-2x)²/2! + (-3)(-4)(-5)(-2x)³/3! = 1 + 6x + 24x² + 80x³. Valid for |2x| < 1, i.e., |x| < 1/2.

Question 2: (8 + 3x)^1/3 = 2(1 + 3x/8)^1/3. Expand with n = 1/3, replacing x with 3x/8: 2[1 + (1/3)(3x/8) + (1/3)(-2/3)(3x/8)²/2! + (1/3)(-2/3)(-5/3)(3x/8)³/3!] = 2[1 + x/8 - x²/64 + 5x³/1536] = 2 + x/4 - x²/32 + 5x³/768.

Question 3: C(8, 4) = 70. The coefficient of x⁴ is 70.

Question 4: 0.98 = 1 - 0.02, so (0.98)^1/2 = (1 + (-0.02))^1/2 = 1 + (1/2)(-0.02) + (1/2)(-1/2)(-0.02)²/2 + ... = 1 - 0.01 - 0.00005 - ... = 0.98995 (5 d.p.).

Question 5: 1/((1 + x)(1 + 2x)) = A/(1 + x) + B/(1 + 2x). A = 1, B = -1 (by cover-up). So (1 + x)^-1 - (1 + 2x)^-1 = (1 - x + x²) - (1 - 2x + 4x²) = x - 3x². Wait, checking: A/(1+x) + B/(1+2x) = 1/((1+x)(1+2x)). Setting x = -1: A/0 is undefined, use cover-up: x = -1 gives A = 1/(1+2(-1)) = 1/(-1) = -1. x = -1/2 gives B = 1/(1+(-1/2)) = 1/(1/2) = 2. So -1/(1+x) + 2/(1+2x). Expand: -(1 - x + x²) + 2(1 - 2x + 4x²) = -1 + x - x² + 2 - 4x + 8x² = 1 - 3x + 7x².

Self-check questions

1. Differentiate y = 3x⁴ - 2x³ + 5x - 7.
2. Find the gradient of the curve y = x³ - 6x² + 9x + 1 at the point where x = 2.
3. A particle moves along a straight line so that its displacement s metres from a fixed point after t seconds is given by s = t³ - 6t² + 9t. Find expressions for the velocity and acceleration, and determine when the particle is momentarily at rest.
4. Find the coordinates of the stationary points on the curve y = 2x³ - 9x² + 12x - 4 and determine their nature.
5. The curve y = x² - 4x + 7 has a minimum point. Find its coordinates and verify that it is indeed a minimum by considering the second derivative.

The igcse 4PM1 the binomial series section rewards careful algebraic manipulation and, above all, correct validity statements. These edexcel igcse further pure mathematics notes give you the core methods, with every concept explained step by step. For the binomial series edexcel igcse exam preparation, including edexcel igcse further pure mathematics revision notes and additional practice, explore the Green Bridge CBT platform to build the fluency the exam demands.