A-Level Maths binomial expansion: positive powers and the general binomial series

Where binomial expansion sits in each specification

The key ideas

For positive integer n, (a + b)^n = a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2 + ... + b^n. The powers of a fall by one each term and the powers of b rise by one, always adding to n. The coefficient nCr = n!/(r!(n - r)!) can be read from Pascal's triangle or a calculator. To find one specific term, you only need the general pattern nCr a^(n-r) b^r.

For other n, the series only works with a 1 in front: (1 + x)^n = 1 + nx + n(n - 1)x^2/2! + n(n - 1)(n - 2)x^3/3! + ... and it never stops. So first take out a factor: (4 + x)^(1/2) = 4^(1/2)(1 + x/4)^(1/2) = 2(1 + x/4)^(1/2). Then replace x by x/4 everywhere, including inside the powers.

The infinite series converges only when the 'x' part has size less than 1. For (1 + x/4)^(1/2) that means |x/4| < 1, so |x| < 4. Questions often ask you to state this range or to use the expansion to estimate a value such as sqrt(4.1), which only works when the substituted x is inside the range.

Worked example 1: the coefficient of x^3 in (2 + 3x)^5

1. The general term is 5Cr × 2^(5 - r) × (3x)^r. For x^3 take r = 3.
2. 5C3 = 10, 2^2 = 4 and (3x)^3 = 27x^3.
3. Multiply: 10 × 4 × 27 = 1080.
4. Answer: the coefficient is 1080. A common slip is to use 3 instead of 3^3 = 27.

Worked example 2 (Cambridge P3, Edexcel P4): expand sqrt(4 + x) up to x^3 and estimate sqrt(4.1)

1. Factor out 4: sqrt(4 + x) = 2(1 + x/4)^(1/2).
2. Use the series with n = 1/2 and x replaced by x/4: 1 + (1/2)(x/4) + (1/2)(-1/2)/2 × (x/4)^2 + (1/2)(-1/2)(-3/2)/6 × (x/4)^3.
3. Simplify each term: 1 + x/8 - x^2/128 + x^3/1024.
4. Multiply by 2: sqrt(4 + x) ≈ 2 + x/4 - x^2/64 + x^3/512, valid for |x| < 4.
5. Put x = 0.1: 2 + 0.025 - 0.00015625 + 0.00000195 = 2.0248457. The true value of sqrt(4.1) is 2.0248457 to 7 decimal places.

Worked example 3: the first three terms of (1 + x)/(1 - 2x)

1. Write it as (1 + x)(1 - 2x)^(-1).
2. (1 - 2x)^(-1) = 1 + (-1)(-2x) + (-1)(-2)/2 × (-2x)^2 + ... = 1 + 2x + 4x^2 + ...
3. Multiply by (1 + x) and keep terms up to x^2: 1 + 2x + 4x^2 + x + 2x^2 = 1 + 3x + 6x^2.
4. Valid for |2x| < 1, that is |x| < 1/2.

Common mistakes that cost marks

• Writing (3x)^3 as 3x^3: the bracket means the 3 is cubed too.
• Losing the sign of a negative second term: (2 - x)^5 has alternating signs.
• Using the infinite series without first making the first term 1.
• Taking out the factor wrongly, for example (4 + x)^(1/2) = 4(1 + x/4)^(1/2) instead of 2(1 + x/4)^(1/2).
• Stating validity for x instead of for the substituted expression, such as |x| < 1 instead of |x| < 4.
• Simplifying fraction coefficients incorrectly when n is negative or fractional.

Exam technique and how a tutor helps

Write each term unsimplified first, with brackets around the whole x-term, and only then simplify. This makes sign errors easy to spot and secures the method mark. When asked to use your expansion to estimate a value, show the substitution and compare with a calculator value only if asked.

A one-to-one tutor usually finds that binomial errors come from algebraic habits, such as powers of a bracketed term or fractions with negatives. Lessons drill those habits with short, targeted questions, then move to past-paper questions where the expansion is combined with partial fractions or an approximation.

Self-check: can you do these?

• Expand (1 + x)^(-2) up to x^3. (Answer: 1 - 2x + 3x^2 - 4x^3)
• Find the term in x^2 in (3 - x)^6. (Answer: 1215x^2)
• For what values of x is the expansion of (2 - 3x)^(-1) valid? (Answer: |x| < 2/3)
• Find 7C3. (Answer: 35)
• Expand (1 - x)^(1/2) up to x^2. (Answer: 1 - x/2 - x^2/8)