Reach one term without expanding: To find one specific term without expanding everything, use the general term: the (r + 1)th term of (a + b)ⁿ is nCr aⁿ⁻ʳ bʳ.
IB-style question — find a specific term
Find the term in x³ in the expansion of (2 + x)⁶.
Step by step
- General term: ⁶Cᵣ 2⁶⁻ʳ xʳ. The power of x is r, so for x³ take r = 3.
- Compute the coefficient.
Final answer
160x³.
Match the power: Decide which power of x you need, set the exponent equal to it to find r, then compute that one coefficient — no full expansion required.
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Pick out one coefficient: A very common question asks for one coefficient — set up the general term, choose the r that gives the power you want, and compute it (watching signs and coefficients).
IB-style question — with a negative term
Find the coefficient of x⁴ in the expansion of (2x − 3)⁶.
Step by step
- General term: ⁶Cᵣ (2x)⁶⁻ʳ (−3)ʳ. The power of x is 6 − r, so for x⁴ take r = 2.
- Compute — square the 2 and the −3.
Final answer
Coefficient = 2160.
IB-style question — when the answer is negative
Find the coefficient of x³ in the expansion of (3x − 2)⁴.
Step by step
- General term: ⁴Cᵣ (3x)⁴⁻ʳ (−2)ʳ. For x³ take r = 1 (the power of x is 4 − 1 = 3).
- An ODD power of −2 keeps the minus, so the coefficient comes out negative.
Final answer
Coefficient = −216. (Odd power of the negative term → negative answer; even power → positive.)
IB-style question — the term independent of x
Find the term independent of x in the expansion of (x + 2/x)⁶.
Step by step
- General term: ⁶Cᵣ x⁶⁻ʳ (2/x)ʳ = ⁶Cᵣ 2ʳ x⁶⁻²ʳ.
- "Independent of x" means the power is 0: 6 − 2r = 0, so r = 3.
- Compute that term.
Final answer
160.
Watch the whole term: Square the whole term: (2x)⁴ = 16x⁴, (−3)² = +9.
And "term independent of x" / "constant term" means the power of x is 0 — solve for r.
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Given a coefficient, solve for the constant: When the exam gives a coefficient and asks for an unknown constant, write that coefficient using the general term, set it equal to the value, and solve.
IB-style question — find k
In the expansion of (x + k)⁷, the coefficient of x⁵ is 84.
Find the possible values of k.
Step by step
- The x⁵ term: power of x is 7 − r = 5, so r = 2.
- Set the coefficient equal to 84.
- Solve (both signs).
Final answer
k = ±2.
IB-style question — a coefficient inside
In the expansion of (x + 2a)⁶, the coefficient of x⁴ is 60.
Find the possible values of a.
Step by step
- The x⁴ term: 6 − r = 4, so r = 2; the second term is 2a.
- Set equal to 60 and solve.
Final answer
a = ±1.
IB-style question — a coefficient on the x-term too
In the expansion of (2x + k)⁶, the coefficient of x⁴ is 2160.
Find the possible values of k.
Step by step
- The x⁴ term: power of x is 6 − r = 4, so r = 2. Raise the WHOLE 2x to its power.
- Set the coefficient equal to 2160 and solve.
Final answer
k = ±3. (The easy slip: forgetting to raise the 2 — (2x)⁴ = 16x⁴, not 2x⁴.)
± or not?: An even power of the unknown (like k²) gives two values (±).
Check whether the question restricts it (e.g. "k > 0" or "the constant is positive").
When the power is unknown: Sometimes n is unknown — use the simplest coefficient (usually x¹ or x²) to form an equation in n, and solve for the positive integer.
IB-style question — from one coefficient
In the expansion of (1 + x)ⁿ, the coefficient of x² is 28.
Find n.
Step by step
- The x² coefficient is ⁿC₂.
- Form and solve the quadratic.
- Take the positive integer.
Final answer
n = 8.
IB-style question — first terms give n and k
The expansion of (1 + kx)ⁿ begins 1 + 12x + 60x² + … .
Find n and k.
Step by step
- First two coefficients: ⁿC₁ k = nk and ⁿC₂ k².
- From the first, k = 12/n; substitute into the second.
- Solve for n, then k.
Final answer
n = 6, k = 2.
IB-style question — a power of x inside the bracket
In the expansion of (2 + x²)ⁿ, the coefficient of x⁴ is 240.
Find n.
Step by step
- The variable term is x², so x⁴ comes from (x²)² — that is r = 2 (not 4).
- Set the coefficient equal to 240 and find the positive integer n.
Final answer
n = 6. (Watch the inside power: x⁴ needs (x²)², so r = 2.)
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Two unknowns ⇒ two equations: The hardest version gives two conditions (two coefficients, or two related expansions) — set up two equations and solve them simultaneously.
IB-style question — two coefficients
In the expansion of (ax + b)⁴, where a, b > 0, the coefficient of x³ is 108 and the coefficient of x² is 54.
Find a and b.
Step by step
- x³ term (r = 1) and x² term (r = 2).
- Tidy: a³b = 27 and a²b² = 9. Divide the first by the second.
- Substitute a = 3b into a³b = 27.
Final answer
a = 3, b = 1.
Divide to eliminate: With two equations in a and b, dividing one by the other usually cancels a variable and leaves a simple equation — much faster than substitution from scratch.