For negative or fractional powers, the binomial expansion of (1 + x)ⁿ becomes an infinite series: 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …, valid only when |x| < 1. Replace x with bx for (1 + bx)ⁿ, which is valid for |x| < 1/|b|. For (a + bx)ⁿ, take out aⁿ first. The expansions are used to approximate roots and to expand fractions after splitting them into partial fractions.
The binomial expansion for rational n
- (1 + x)ⁿ = 1 + nx + n(n − 1)/2! x² + n(n − 1)(n − 2)/3! x³ + …, valid for |x| < 1
- (1 + bx)ⁿ: valid for |bx| < 1, that is |x| < 1/|b|
- (a + bx)ⁿ = aⁿ(1 + (b/a)x)ⁿ, valid for |x| < |a/b|
- Percentage error = |estimate − true| / true × 100%
Expanding (1 + x)ⁿ for any rational n
When n is negative or a fraction, the expansion never stops: it is an infinite series. Use the same pattern of coefficients, building each one carefully: the x³ coefficient is n(n − 1)(n − 2)/3!, keeping all three factors. Give coefficients as fractions in their simplest form, and always state the range of validity, |x| < 1.
Expanding (1 + bx)ⁿ and its validity
Replace x with bx everywhere, putting the whole of bx in brackets at every power: (−4x)³ = −64x³. The expansion is valid when |bx| < 1, which gives |x| < 1/|b|. For example, √(1 − 4x) is valid for |x| < ¼.
Expanding (a + bx)ⁿ
If the bracket does not start with 1, take out the first number before expanding: (a + bx)ⁿ = aⁿ(1 + (b/a)x)ⁿ. Then expand the bracket and multiply every term by aⁿ. Remember that aⁿ may be a root: 4^(1/2) = 2 and 8^(1/3) = 2. Forgetting to raise a to the power n is the most common error. The validity becomes |x| < |a/b|.
Approximations and percentage error
To approximate a root, choose a small value of x so that the bracket becomes the number you want, check that it lies within the range of validity, and substitute into the first few terms. The percentage error compares the estimate with the calculator value: |estimate − true| / true × 100%.
Expanding fractions using partial fractions
To expand a fraction such as (3 + 2x)/((1 − x)(1 + 2x)), first split it into partial fractions, then write each as a negative power, expand each one, and add the matching terms. If the fraction is improper, divide first. The combined expansion is valid only where every part is valid, so take the narrower range.
A factor that multiplies a binomial
If a function is a polynomial factor times a binomial, such as (3 − 2x)√(1 + x), expand only the binomial, then multiply by the factor and collect like powers. The x² term comes from the constant times the x² term plus the x term times the x term.
Worked examples
Example 1. Find the first four terms, in ascending powers of x, of √(1 + x).
- n = ½. x coefficient: ½.
- x² coefficient: (½)(−½)/2 = −1/8.
- x³ coefficient: (½)(−½)(−3/2)/6 = 1/16.
Answer: 1 + ½x − ⅛x² + (1/16)x³, valid for |x| < 1.
Marks: M1 for the structure of the x³ coefficient, A1 for −⅛x², A1 for (1/16)x³.
Example 2. Find the first four terms of √(1 − 4x), and state when the expansion is valid.
- Use the coefficients 1, ½, −1/8, 1/16 with u = −4x.
- ½(−4x) = −2x; −⅛(16x²) = −2x²; (1/16)(−64x³) = −4x³.
- Valid when |−4x| < 1.
Answer: 1 − 2x − 2x² − 4x³, valid for |x| < ¼.
Marks: M1 for substituting −4x, A1 for −2x², A1 for −4x³, B1 for the validity.
Example 3. Find the first four terms of √(4 + x), and state when the expansion is valid.
- Take out 4^(1/2) = 2: √(4 + x) = 2(1 + x/4)^(1/2).
- (1 + x/4)^(1/2) = 1 + x/8 − x²/128 + x³/1024 − …
- Multiply every term by 2.
Answer: 2 + ¼x − (1/64)x² + (1/512)x³, valid for |x| < 4.
Marks: M1 for taking out 2, A1 for each coefficient, B1 for the validity.
Example 4. Using √(1 − 2x) ≈ 1 − x − ½x², estimate √0.98 by substituting x = 0.01.
- 1 − 2(0.01) = 0.98, so x = 0.01 gives √0.98, and 0.01 is within |x| < ½.
- 1 − 0.01 − ½(0.0001) = 1 − 0.01 − 0.00005.
Answer: √0.98 ≈ 0.98995.
Marks: M1 for substituting x = 0.01, A1 for 0.98995.
Example 5. Express (3 + 2x)/((1 − x)(1 + 2x)) in partial fractions, then find the first three terms of its expansion.
- 3 + 2x = A(1 + 2x) + B(1 − x). x = 1 gives A = 5/3; x = −½ gives B = 4/3.
- (1 − x)⁻¹ = 1 + x + x² + … and (1 + 2x)⁻¹ = 1 − 2x + 4x² − …
- Combine: (5/3)(1 + x + x²) + (4/3)(1 − 2x + 4x²).
Answer: 3 − x + 7x², valid for |x| < ½.
Marks: M1 for the partial fractions, A1 for A and B, M1 for expanding each, A1 for 3 − x + 7x².
Example 6. Find the first three terms of (3 − 2x)√(1 + x).
- √(1 + x) = 1 + ½x − ⅛x² + …
- Constant: 3. x term: 3(½) + (−2)(1) = −½.
- x² term: 3(−⅛) + (−2)(½) = −3/8 − 1 = −11/8.
Answer: 3 − ½x − (11/8)x².
Marks: M1 for the binomial, M1 for multiplying and collecting, A1 for −(11/8)x².
Common mistakes
- Not stating the validity. Always give the range, such as |x| < ¼; it often carries a mark.
- Forgetting to take out aⁿ. (4 + x)^(1/2) = 2(1 + x/4)^(1/2), not (1 + x/4)^(1/2).
- Dropping a factor in the x³ coefficient. It is n(n − 1)(n − 2)/3!, with all three factors.
- Not bracketing bx. (−4x)² = 16x², not −16x².
- Using a value outside the valid range. An approximation only works if x is within the range of validity.
How the general binomial expansion is examined
Negative and fractional powers move the binomial expansion into Pure 4 (WMA14) for Edexcel International A-Level; 9MA0 covers the same material in its second year. Questions usually ask you to:
- find the first three or four terms of (a + bx)ⁿ, with coefficients as simplified fractions
- state the range of values of x for which the expansion is valid
- use the expansion to estimate a root, sometimes with a percentage error
- expand a fraction after splitting it into partial fractions
Find full questions with mark schemes in the Pure 4 binomial expansion exam questions. The Pure 4 binomial expansion worksheet and the sequences and series revision engine give more practice.
This extends the binomial expansion for positive whole n, and combines with partial fractions.
Frequently asked questions
When is the binomial expansion of (1 + x)ⁿ valid?
For a positive whole number n, it is valid for all x. For negative or fractional n, it is valid only when |x| < 1.
How do you expand (a + bx)ⁿ when a is not 1?
Take out aⁿ: (a + bx)ⁿ = aⁿ(1 + (b/a)x)ⁿ, then expand the bracket and multiply every term by aⁿ.
Why does the expansion have infinitely many terms for negative n?
Because the coefficients n(n − 1)(n − 2)… never reach a factor of zero when n is negative or a fraction.