Maths › Sequences and series › The binomial expansion
The binomial expansion
Raising a bracket to the tenth power by repeated multiplication is an afternoon gone. The binomial theorem does it in one line, with coefficients that have been sitting in Pascal's triangle all along, and for small x the first few terms do nearly all the work of the whole expansion.
Builds on Indices and surds.
IN THIS TOPIC
- Compute binomial coefficients from Pascal's triangle or the nCr formula.
- Expand (a + bx)n for positive integer n, and pick out single coefficients.
- Truncate an expansion to approximate powers of numbers close to 1.
WHAT YOU PROBABLY THINK
The coefficient of xr in (a + bx)n is nCr times br.
Pascal's triangle and nCr
Expanding (a + b)n means choosing, from each of the n brackets, either an a or a b, and collecting the results. The number of ways of choosing exactly r of the brackets to donate a b is the binomial coefficient
where n! means n × (n − 1) × … × 1. The same numbers fill Pascal's triangle, in which every entry is the sum of the two above it, and both notations appear on papers, nCr and the two-row bracket alike.
WORKED EXAMPLE
A coefficient three ways
Evaluate 10C3.
By formula: 10!/(3! × 7!) = (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120.
The factorials mostly cancel; only three factors of the top survive. A calculator's nCr button agrees, and row 10 of Pascal's triangle would too, eventually.
For hand work, cancel before multiplying. Computing 10! in full is how calculators spend their time, not how you should spend yours.
The theorem
The full statement, printed in the formulae booklet, is
and its shape matters more than its symbols. Powers of a fall as powers of b rise, every term has total degree n, and the coefficients read straight from row n of the triangle. This is where the opening lie dies. Each term keeps its an−r factor, so with b = 3x and a = 2 the x-coefficient of (2 + 3x)4 is 4C1 × 23 × 3 = 96, a long way from the bare 4 × 3 = 12.
WORKED EXAMPLE
A full expansion, kept tidy
Expand (2 + 3x)4 completely.
Row 4 gives coefficients 1, 4, 6, 4, 1. Build each term as (coefficient) × 24−r × (3x)r.
The terms are 16, then 4 × 8 × 3x = 96x, then 6 × 4 × 9x2 = 216x2, then 4 × 2 × 27x3 = 216x3, then 81x4.
So (2 + 3x)4 = 16 + 96x + 216x2 + 216x3 + 81x4.
A substitution check is available and cheap. At x = 1 the bracket is 54 = 625, and the terms sum 16 + 96 + 216 + 216 + 81 = 625.
YOUR TURN
One coefficient, minus sign included
Find the coefficient of x3 in the expansion of (3 − 2x)5, before opening the working.
Show the working
The x3 term is 5C3 × 32 × (−2x)3 = 10 × 9 × (−8)x3.
The coefficient is −720.
The minus sign lives inside the bracket, so it gets cubed along with the 2. Leaving it outside the power is the most common lost mark in this topic.
Expansions as approximations
When x is small, x2 is smaller and x3 smaller still, so the early terms of an expansion carry nearly all of its value. Truncating after two or three terms turns the theorem into an approximation machine for powers of numbers near 1.
WORKED EXAMPLE
Estimating a tenth power by hand
Expand (1 + x/2)10 up to the x2 term, and use the result to estimate 1.00510.
The terms are 1, then 10 × (x/2) = 5x, then 45 × (x/2)2 = 11.25x2.
Setting x/2 = 0.005 means x = 0.01, and the truncation gives 1 + 0.05 + 0.001125 = 1.051125.
The true value is 1.0511401…, so the estimate is inside 0.00002. The x3 term would have contributed about 0.000015, which is the error almost to the digit.
TRY IT UNSEEN
A quick estimate of 1.02⁶
Expand (1 + 2x)6 in ascending powers of x up to the x2 term, and use it to estimate 1.026.
Show the working
The terms are 1, then 6 × 2x = 12x, then 15 × 4x2 = 60x2.
For 1.02 the substitution is 2x = 0.02, so x = 0.01, giving 1 + 0.12 + 0.006 = 1.126.
The true value is 1.12616…, and the shortfall is the tail the truncation threw away. Quote estimates to the accuracy the truncation supports, not to nine calculator digits.
THE EXAM BIT
- State which row of coefficients you are using, then build terms as coefficient × a-power × (bx)-power. Marks attach to visible structure.
- Keep the whole of bx inside the power, sign included: (−2x)3 is −8x3, and dropping that minus is the classic error here.
- “In ascending powers of x” fixes the order of your answer; start from the constant term.
- Check a full expansion by substituting x = 1: the bracket evaluates in one step and must equal the sum of your terms.
- For approximation parts, identify x from the number given, state the substitution, and round only at the end.
CHECK YOURSELF
Find the first three terms, in ascending powers of x, of (2 − x)5, and use them to estimate 1.9985.
Show a hint
a = 2, b = −1; then 2 − x = 1.998 fixes x.
Show the answer
The terms are 25 = 32, then 5 × 24 × (−x) = −80x, then 10 × 23 × x2 = 80x2.
So (2 − x)5 ≈ 32 − 80x + 80x2, and 1.998 means x = 0.002.
The estimate is 32 − 0.16 + 0.00032 = 31.84032, and the true value 31.840319… agrees to every digit shown.
Row n of Pascal's triangle is nCr, and every term of (a + b)ⁿ keeps total degree n.
For small x the early terms are the whole story; truncate and say what you dropped.
CHECK YOUR PROGRESS
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- Compute binomial coefficients from Pascal's triangle or the nCr formula.
- Expand (a + bx)n for positive integer n, and pick out single coefficients.
- Truncate an expansion to approximate powers of numbers close to 1.
No animated video for this topic yet; these notes stand alone.