From whole powers to any rational power
You already know (1 + x)^n when n is a positive whole number. The coefficients come from nCr, and the series has n + 1 terms.
The same pattern works for any rational n (negative or a fraction):
(1 + x)^n = 1 + nx + n(n−1)/2! x² + n(n−1)(n−2)/3! x³ + …
Why does it never stop? A coefficient becomes zero only when one of the brackets (n − k) is zero. That only happens if n is a whole number 0, 1, 2, … For n = −1 or n = ½ no bracket is ever zero, so there are infinitely many terms.
Example of coefficients
n = −1: 1, −1, 1, −1, … so (1 + x)⁻¹ = 1 − x + x² − x³ + …
n = ½: 1, ½, −⅛, 1/16, …
When is the expansion valid?
An endless sum only has a value if its terms shrink fast enough. For the binomial series that happens exactly when |x| < 1, that is −1 < x < 1.
Test it: put x = 2 in 1 − x + x² − … You get 1 − 2 + 4 − 8 + … which jumps about wildly. But 1/(1 + 2) = ⅓. The series is simply wrong there.
Always write the validity next to your answer, for example: (1 + 3x)^−½ is valid for |3x| < 1, so |x| < ⅓.
Expanding (a + bx)^n
The formula needs a 1 at the front. So factor out a:
(a + bx)^n = a^n (1 + bx/a)^n
Now expand with x replaced by bx/a, and multiply every term by a^n. The validity is |bx/a| < 1, so |x| < |a/b|.
Example: (4 − x)^½ = 2(1 − x/4)^½, valid for |x| < 4.
With partial fractions
To expand something like (5x + 1)/((1 + x)(1 − 2x)), split it into partial fractions first, expand each part, and add. The whole thing is valid only where both parts are valid, so take the smaller range.
Approximations and exam technique
When x is small, x², x³ … are tiny, so a few terms give an excellent answer. Choose x so that the bracket matches the number you want, e.g. √(4.04) = (4 + 0.04)^½ = 2(1 + 0.01)^½.
Exam questions usually ask: expand up to x² or x³; state the range of validity; use the expansion to estimate a number; find the percentage error; or find unknown constants from given coefficients.
Try it
On the 3D slider set n = −1 and x = 0.9, then x = 1.1. In the first case the green sum creeps towards 1/1.9 ≈ 0.526. In the second the bars grow and the sum flips up and down. Then work out 1/0.98 on paper using (1 − 0.02)⁻¹ ≈ 1 + 0.02 + 0.0004 and check with a calculator.
Key formulas and definitions
- (1 + x)^n = 1 + nx + n(n−1)x²/2! + n(n−1)(n−2)x³/3! + …, valid for |x| < 1 (any rational n)
- General term: n(n−1)…(n−r+1)/r! · x^r
- (a + bx)^n = a^n(1 + bx/a)^n, valid for |bx/a| < 1
- (1 + x)⁻¹ = 1 − x + x² − x³ + …
- (1 − x)⁻¹ = 1 + x + x² + x³ + …
- Small x: (1 + x)^n ≈ 1 + nx
Worked examples
1. Expand (1 + x)⁻² up to x³.
n = −2. Terms: 1 + (−2)x + (−2)(−3)/2 x² + (−2)(−3)(−4)/6 x³ = 1 − 2x + 3x² − 4x³. Valid for |x| < 1.
2. Expand (1 − 3x)^½ up to x² and state when it is valid.
Put y = −3x and n = ½: 1 + ½y − ⅛y². So 1 + ½(−3x) − ⅛(9x²) = 1 − 1.5x − 1.125x². Valid for |3x| < 1, so |x| < ⅓.
3. Expand (2 + x)⁻¹ up to x².
(2 + x)⁻¹ = 2⁻¹(1 + x/2)⁻¹ = ½(1 − x/2 + x²/4) = ½ − x/4 + x²/8. Valid for |x| < 2.
4. Use (1 + x)^½ to estimate √1.02 to 5 decimal places.
x = 0.02: 1 + ½(0.02) − ⅛(0.0004) = 1 + 0.01 − 0.00005 = 1.00995. A calculator gives 1.0099505, so the error is tiny.
5. Expand (8 + 3x)^⅓ up to x².
8^⅓(1 + 3x/8)^⅓ = 2[1 + ⅓(3x/8) + (⅓)(−⅔)/2 (3x/8)²] = 2[1 + x/8 − (1/9)(9x²/64)] = 2 + x/4 − x²/32. Valid for |x| < 8/3.
6. Split (3 + 2x)/((1 + x)(1 + 2x)) into partial fractions and expand up to x².
(3 + 2x)/((1 + x)(1 + 2x)) = A/(1 + x) + B/(1 + 2x). x = −1: 1 = A(−1) so A = −1. x = −½: 2 = B(½) so B = 4. Expand: −(1 − x + x²) + 4(1 − 2x + 4x²) = 3 − 7x + 15x². Valid for |x| < ½ (the smaller range).
Common mistakes
- Forgetting to take a^n outside: (4 + x)^½ is 2(1 + x/4)^½, not 4(1 + x/4)^½ and not (1 + x/4)^½.
- Not putting brackets round bx: in (1 − 3x)^½ the x² term uses (−3x)² = 9x², not −3x².
- Writing no range of validity, or giving |x| < 1 when it should be |bx/a| < 1.
- Thinking the series stops for negative or fractional n. Only positive whole n gives a finite series.