SO / EXAM STUDIO

Mathematics HL / 1 · Number and algebra

04 Binomial expansions

20-minute guided session≈ 8 min reading incl. reveals37 bank questions

Start with the exam

What you need to be able to do

In coefficient questions, the main task is choosing the correct term, not expanding everything. Identify the power of x contributed by each choice and solve its exponent equation first. The bank includes negative powers, outside factors, products of expansions and unknown integer exponents. Separate “coefficient” from “term”: a term includes its power of x. For non-integer powers, a finite truncation is an approximation with a convergence restriction.

Your 20-minute study route

Spend 3 minutes reading the overview and diagram, 7 minutes learning the question patterns, 5 minutes writing out the worked example, and 5 minutes on a bank question or recall check. Take longer when you need to; the full bank is here for later practice.

x²x²−2/x−2/x−2/x−2/xConstant term: select two x² and four −2/xx⁴ × x⁻⁴ = 1 • C(6,4) selections
The diagram shows one selection; the binomial coefficient counts all positions for the four negative-power choices.

Recognise → set up → write

How the questions are set

QUESTION PATTERN 01

Target a power or constant term

Recognise it: Find the coefficient of x^k, the term independent of x, or a specific term.

  1. Write T_(r+1)=C(n,r)A^(n−r)B^r. Track powers of x separately from the numerical coefficient.
  2. Solve the exponent equation for integer r with 0≤r≤n. Include any power of x outside the bracket before setting the target exponent.
  3. Substitute r only after checking it is allowed; if no valid integer r exists, that power is absent.
What to put on the page

For (x²−2/x)^n, show exponent 2(n−r)−r=2n−3r. The minus sign belongs in (−2)^r.

Common mark loss

The index r is one less than the term number. A constant term means exponent zero, not coefficient zero.

QUESTION PATTERN 02

Compare unknown coefficients

Recognise it: Two coefficients are equal, in a ratio, or form a progression.

  1. Extract only the needed coefficients, keeping n and any constant parameter symbolic.
  2. Translate the stated relationship accurately: arithmetic means 2b=a+c; geometric means b²=ac when applicable.
  3. Solve and check that n is a permitted nonnegative integer and that any stated sign or nonzero conditions hold.
What to put on the page

Write separate coefficient expressions before relating them; this shows why a binomial coefficient such as n(n−1)/2 appears.

Common mark loss

Do not compare entire terms when the question compares coefficients. Do not drop the zero-parameter case without checking it.

QUESTION PATTERN 03

Products and approximations

Recognise it: Two expansions are multiplied, or a nearby numerical value is estimated.

  1. To obtain x^k in a product, list all exponent pairs that sum to k; add their coefficient products.
  2. For (1+u)^α use 1+αu+α(α−1)u²/2+… and keep only the requested order. State |u|<1 for the usual non-integer expansion.
  3. Choose a small u representing the requested number, retain guard digits and distinguish the truncated approximation from an exact finite expansion.
What to put on the page

Write the two or three contributing pairs explicitly; this is safer than multiplying long polynomials and searching afterwards.

Common mark loss

For an infinite expansion, “up to x²” does not mean all higher terms vanish. A large |u| can invalidate the approximation.

Write it before you reveal it

A worked exam-style example

Original teaching example, created for this lesson.

Original practice: find the constant term in (x²−2/x)^6.

Reveal the worked solution
  1. The general term is C(6,r)(x²)^(6−r)(−2/x)^r.
  2. Its power is x^(12−3r), so set 12−3r=0 and obtain r=4.
  3. The coefficient is C(6,4)(−2)^4=15×16.

Answer: The constant term is 240.

Check: Four choices of −2/x give x^(−4), and two choices of x² give x^4; the powers cancel and the sign is positive.

From your own collection

Apply it to past-paper questions

These questions anchor the patterns above. Attempt one without the answer, then compare your setup and conclusion with the reasoning. The linked reader contains the source answer and the question-specific syllabus review.

Q141 · 2022 · PAPER 1 · 5 MARKS

Question 141

Apply the method to the original question.

Compatibility: This current-course AA HL question tests the listed current syllabus methods.

View the original question

Original page images preserve the equations and diagrams. Write your response before revealing the reasoning.

Original Mathematics question 141, page 253
Reveal the revision reasoning

Selecting the second term r times gives a nonzero multiple of x^[3(n-r)-r]=x^(3n-4r).

A constant term requires 3n-4r=0, so r=3n/4.

Since r is an integer and 3 and 4 are coprime, n must be a positive multiple of 4.

Conversely n=4m gives r=3m within 0≤r≤n, so every n=4,8,12,… works.

Editorial revision reasoning. Check the original answer for its marking detail.

Q140 · 2023 · PAPER 1 · 6 MARKS

Question 140

Apply the method to the original question.

Compatibility: All parts test current AA HL content and can be solved using current syllabus methods.

View the original question

Original page images preserve the equations and diagrams. Write your response before revealing the reasoning.

Original Mathematics question 140, page 252
Reveal the revision reasoning

The binomial theorem gives coefficient nk for x and n(n-1)k²/2 for x².

The x coefficient gives nk=12, so k is nonzero. Comparing x² coefficients then allows cancellation of k², giving n(n-1)/2=28.

Solve n²-n-56=(n-8)(n+7)=0. Since n is a positive integer, n=8.

Return to nk=12 to obtain k=3/2.

Editorial revision reasoning. Check the original answer for its marking detail.

Q143 · 2021 · PAPER 1 · 7 MARKS

Question 143

Apply the method to the original question.

Compatibility: This AA HL question uses current syllabus methods in the references shown.

View the original question

Original page images preserve the equations and diagrams. Write your response before revealing the reasoning.

Original Mathematics question 143, page 255
Reveal the revision reasoning

The expansions are (1+ax)^(-1/2)=1-(a/2)x+(3a²/8)x²+… and (1-x)^(1/2)=1-x/2-x²/8+….

Their difference has coefficients (1-a)/2 for x and (3a²+1)/8 for x².

Match these to 4b and b. Eliminating b gives 1-a=3a²+1, so a(3a+1)=0.

Since a≠0, a=-1/3 and b=1/6.

The two series require abs(ax)<1 and abs(x)<1; together, with this a, they give -1<x<1.

Editorial revision reasoning. Check the original answer for its marking detail.

All 37 questions in this topic

Close your notes first

Can you retrieve the method?

1. What r gives the third term?

Check answer 1

r=2.

2. Which terms make x³ in a product?

Check answer 2

All available exponent pairs summing to 3, such as (0,3),(1,2),(2,1),(3,0).

3. Why must the exponent equation be solved first?

Check answer 3

It selects the correct term and can show that the requested power is absent.

Before you move on

  1. For each coefficient answer, display the binomial factor, constant powers and exponent selection. Those are the meaningful method steps.
  2. Check the sign using the parity of r and check whether the requested answer is a term, a coefficient, or an approximation.

Your revision note

Record one missed cue, one corrected step and one question to retry. Saved only in this browser.