Most tested MM2.4

The Binomial Expansion

The Binomial Theorem provides a powerful and quick method for expanding brackets raised to a positive integer power, of the form (a + b)n. It allows you to find any specific term or coefficient without the lengthy process of manual multiplication.

Part of the ESAT Mathematics 2 syllabus — revision for the Engineering and Science Admissions Test (ESAT), the UAT-UK admissions test for Cambridge, Imperial, Oxford and UCL.

Key points

  • The expansion of (a + b)n contains n + 1 terms.
  • For each term in the expansion, the sum of the powers of 'a' and 'b' is always equal to n.
  • The coefficient of any term is determined by the binomial coefficient, (n choose r), which can be calculated using factorials.
  • When expanding an expression like (a + f(x))n, ensure the power is applied to the entirety of f(x), including its sign and any numerical coefficient.
  • The binomial coefficients are symmetrical, meaning (n choose r) is always equal to (n choose n-r). This can simplify calculations.

Formulae

(a + b)n = (n choose 0)an + (n choose 1)a^(n-1)b + ⋯ + (n choose r)a^(n-r)b^r + ⋯ + (n choose n)bn

To expand any expression of the form (a + b)n for a positive integer n.

The general term is (n choose r) × a^(n-r) × b^r

To find a single, specific term in the expansion without calculating the whole series. The term is the (r+1)th term.

Definitions

n! (n factorial)
The product of all positive integers up to n. For example, 5! = 5 × 4 × 3 × 2 × 1 = 120. By convention, 0! = 1.
(n choose r)
The binomial coefficient, representing the number of ways to choose r items from a set of n. It is calculated as n! / (r! × (n-r)!).

Worked example

Find the constant term (the term independent of x) in the expansion of (2x2 - 1/x)6.

  1. 1

    First, write down the general term of the expansion using the formula (n choose r) × a^(n-r) × b^r.

    Here n=6, a=2x2, and b=-1/x
  2. 2

    The general term is (6 choose r) × (2x2)^(6-r) × (-1/x)^r.

  3. 3

    Now, separate the coefficients from the x parts:

    (6 choose r) × 2^(6-r) × (-1)^r × (x2)^(6-r) × (x-1)^r.

  4. 4

    Combine the powers of x:

    x^(2*(6-r)) × x^(-r) = x^(12 - 2r - r) = x^(12 - 3r).

  5. 5

    For the constant term, the power of x must be zero.

    So, set 12 - 3r = 0.

  6. 6

    Solving for r gives 3r = 12, so r = 4.

  7. 7
    Substitute r=4 back into the coefficient part of the general term:

    (6 choose 4) × 2^(6-4) × (-1)4.

  8. 8

    Calculate the value.

    (6 choose 4) is the same as (6 choose 2) = (6*5)/(2*1) = 15.

    The term becomes 15 × 22 × 1 = 15 × 4 = 60.

Answer: 60

Common mistakes

  • ×Forgetting to apply the power to the entire term. In an expansion of (3x)4, the coefficient is 34, not just 3. This is a frequent source of arithmetic errors.
  • ×Sign errors. Remember that a negative term, like (-2y), raised to an odd power will be negative, while raising it to an even power will result in a positive term.
  • ×Miscalculating the combined power of x when terms involve fractions, e.g., getting x^(a-b) wrong when the terms are x^a and 1/x^b.

No-calculator tips

  • ✓Simplify binomial coefficients before multiplying. For (10 choose 3), calculate (10*9*8)/(3*2*1) and cancel terms (e.g., 9/3=3, 8/2=4) to get 10*3*4 = 120, avoiding large factorials.
  • ✓Use the symmetry rule (n choose r) = (n choose n-r) to your advantage. Calculating (12 choose 10) is much harder than calculating (12 choose 2), but they give the same answer.
  • ✓Memorise the first few powers of 2 (up to 210) and 3 (up to 35) as they appear frequently in coefficient calculations.

Read this topic in the official UAT-UK ESAT guide →

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