Key Idea: The binomial theorem expands a power of a two-term bracket, (a + b)ⁿ, without multiplying it all out — and lets you grab one term straight away. It shows up on both papers, often as find the coefficient of xᵏ or find the constant term.
🔢 The formulas you're given
| Pascal's triangle | ⁿCᵣ (combinations) |
|---|---|
| Start each row with 1; every inside number is the sum of the two above. | Each coefficient is ⁿCᵣ — compute by formula or with the GDC's nCr. |
| Row 4: 1 4 6 4 1. | ⁴C₂ = 6, matching the triangle. |
| Quickest for small n (≈ up to 6) — Paper 1. | Best for large n — Paper 2 on the GDC. |
- the power of the bracket (top of ⁿCᵣ)
- which term — 0 for the first, counting up
- first term — its power **falls** n → 0
- second term — its power **rises** 0 → n
(a + b)ⁿ has n + 1 terms, and in each term the two powers sum to n. Need just one term? Use the general term ⁿCᵣ aⁿ⁻ʳ bʳ: set the exponent of x equal to the power you want, solve for r, then compute that single coefficient — no full expansion.
✏️ IB-style worked examples
IB-style question — expand a bracket (Paper 1)
Expand (x + 2)⁴ without a calculator.
Step by step:
Row 4 of Pascal's triangle gives 1, 4, 6, 4, 1; powers of x fall, powers of 2 rise.
Work out each coefficient (2² = 4, 2³ = 8, 2⁴ = 16).
(x + 2)⁴ = x⁴ + 8x³ + 24x² + 32x + 16 (5 terms; powers sum to 4)
IB-style question — a coefficient with the nCr (Paper 2)
Find the coefficient of x³ in the expansion of (2x − 1)⁵.
Step by step:
General term: ⁵Cᵣ (2x)⁵⁻ʳ (−1)ʳ. The power of x is 5 − r, so for x³ take r = 2.
Use the GDC for ⁵C₂ = 10, then cube the whole 2x and square the −1.
Coefficient of x³ = 80
IB-style question — find an unknown constant
In the expansion of (x + k)⁶, the coefficient of x⁴ is 60. Find the possible values of k.
Step by step:
The x⁴ term: power of x is 6 − r = 4, so r = 2.
Set the coefficient equal to 60.
Solve — an even power gives both signs.
k = ±2
🔒 GDC walkthrough
Step through the exact calculator keystrokes, screen by screen, in study mode.
Important: A coefficient or minus sign is part of the term, so raise the whole thing: (2x)³ = 8x³ (not 2x³), (−3)² = +9 (not −9). And "term independent of x" / "constant term" means the power of x is 0 — set the exponent to 0 and solve for r.
Tap each card to reveal the answer.
Coefficients of (a + b)⁴ from Pascal's triangle? 1, 4, 6, 4, 1 — row 4 of the triangle.
Compute ⁵C₂ 10 — 5!/(2!·3!) = (5×4)/(2×1).
How many terms in (a + b)⁷? 8 terms — always n + 1.
Term in x³ of (2 + x)⁶? 160x³ — ⁶C₃·2³·x³ = 20×8×x³.
Constant term of (x + 2/x)⁶? 160 — set 6 − 2r = 0 → r = 3, then ⁶C₃·2³.
(1 + x)ⁿ has x² coefficient 28 — find n n = 8 — ⁿC₂ = n(n−1)/2 = 28 ⇒ n² − n − 56 = 0.
Exam tips
- Coefficients of (a + b)ⁿ = row n of Pascal's triangle = ⁿC₀, ⁿC₁, …, ⁿCₙ.
- Paper 1: small powers by hand. Paper 2: ⁿCᵣ via MATH → ▶ PRB → 3:nCr.
- General term ⁿCᵣ aⁿ⁻ʳ bʳ — match the exponent to the power you want to get r.
- Bracket the whole term before raising: (2x)³ = 8x³, (−3)² = +9; constant term ⇒ power of x is 0.
- Coefficient given → set it equal and solve (even power ⇒ ±); n unknown → use ⁿC₂, a quadratic in n.