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, and the coefficients have been sitting in Pascal's triangle the whole time. For small x the first two or three terms carry almost all the value, which turns the theorem into a way of estimating awkward powers by hand.
Builds on Indices and surds.
IN THIS TOPIC
- Compute binomial coefficients from Pascal's triangle or from the nCr formula.
- Expand (a + bx)n for positive integer n, and pull out one coefficient without expanding the rest.
- Truncate an expansion to estimate a power of a number close to 1.
COMMON MISCONCEPTION
The coefficient of xr in (a + bx)n is nCr times br.
Coefficients, and the theorem
Expanding (a + b)n means walking along the n brackets, taking either an a or a b from each, and collecting like terms at the end. Count the ways of picking r of those brackets to donate a b and you have the binomial coefficient
with n! standing for n × (n − 1) × … × 1. The same numbers fill Pascal's triangle, where every entry is the sum of the two above it. Papers write the coefficient three ways, as nCr, as a two-row bracket, and as nCr, and you are expected to recognise all of them. The identical count reappears in binomial probability on the statistics paper.
WORKED EXAMPLE
A coefficient, done properly
Evaluate 10C3.
The formula gives 10!/(3! × 7!). Nearly everything cancels, leaving (10 × 9 × 8)/(3 × 2 × 1) = 720/6 = 120.
A calculator's nCr button agrees. So does row 10 of Pascal's triangle, eventually.
Cancel before you multiply. Working 10! out in full is how a calculator spends its time, and it is not how you should spend yours.
The full statement of the theorem is printed in the formulae booklet.
Its shape matters more than its symbols. Powers of a fall away as powers of b climb, and every term has total degree n. The coefficients read straight off row n of the triangle.
Each term drags its an−r factor along with it, so in (2 + 3x)4 the coefficient of x is 4C1 × 23 × 3 = 96, not the 4 × 3 = 12 that forgetting the factor would predict.
WORKED EXAMPLE
A full expansion, kept tidy
Expand (2 + 3x)4 completely.
Row 4 gives 1, 4, 6, 4, 1. Build each term as (coefficient) × 24−r × (3x)r.
That produces 16, then 4 × 8 × 3x = 96x, then 6 × 4 × 9x2 = 216x2, then 4 × 2 × 27x3 = 216x3, and finally 81x4.
So (2 + 3x)4 = 16 + 96x + 216x2 + 216x3 + 81x4.
Now check it. At x = 1 the bracket is 54 = 625, and the five terms sum to 16 + 96 + 216 + 216 + 81 = 625. That takes two seconds and catches almost every arithmetic slip available in this topic.
GUIDED PRACTICE
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 here, and it takes the whole answer with it.
Expansions as approximations
When x is small, x2 is smaller and x3 is smaller still. Chop the expansion after two or three terms and you have discarded almost nothing, which is what makes the theorem a tool for estimating powers of numbers near 1 without a calculator.
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.
INDEPENDENT PRACTICE
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 you threw away. Quote estimates to the accuracy the truncation supports, not to nine calculator digits.
ASSESSMENT FOCUS
- Say which row of coefficients you are using, then build terms as coefficient × a-power × (bx)-power. Marks attach to visible structure, not to a bare final number.
- Keep the whole of bx inside the power, sign included. (−2x)3 is −8x3.
- “In ascending powers of x” fixes the order of your answer. Start at the constant term and work up.
- For approximation parts, state what x is, substitute it into the truncation, and round once at the very 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. The true value is 31.840319…, which 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 carry nearly everything; truncate, then say what you dropped.
WORKBOOK
Printable practice for this topic: original exam-style questions with room to work, and a fully worked answer book. Free to use; please do not redistribute or sell.
Or read them with their worked answers on the the binomial expansion questions page.
CHECK YOUR PROGRESS
Rate how confident you feel with each objective for this lesson. Ratings are saved in this browser, on this device, unless you sign in.
- Compute binomial coefficients from Pascal's triangle or from the nCr formula.
- Expand (a + bx)n for positive integer n, and pull out one coefficient without expanding the rest.
- Truncate an expansion to estimate a power of a number close to 1.
Open the full revision checklist to see every objective in the course in one place.