This set mixes every skill from binomial expansion: full expansions, single terms, unknown constants, negative parts and substitution checks. The questions are original and ordered from easier to harder.
Write your working on paper before opening each answer. Keep every part of a term in brackets until you have simplified it, and check each expansion with x = 1 when you can. The non-calculator working trainer is useful for the exact arithmetic.
Questions and worked answers
1. Expand (x + 1)⁴.
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Coefficients 1, 4, 6, 4, 1, and every power of 1 is 1.
x⁴ + 4x³ + 6x² + 4x + 1
Check at x = 1: 2⁴ = 16, and 1 + 4 + 6 + 4 + 1 = 16.
2. Write the row of binomial coefficients for power 7 and check its sum.
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Each number is the sum of the two above it in Pascal’s triangle, starting from the power 6 row (1, 6, 15, 20, 15, 6, 1).
1, 7, 21, 35, 35, 21, 7, 1
The sum is 128 = 2⁷.
3. Expand (2x + 1)⁴.
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Coefficients 1, 4, 6, 4, 1. Terms: (2x)⁴ = 16x⁴; 4 × (2x)³ = 4 × 8x³ = 32x³; 6 × (2x)² = 6 × 4x² = 24x²; 4 × 2x = 8x; 1.
16x⁴ + 32x³ + 24x² + 8x + 1
Check at x = 1: 3⁴ = 81, and 16 + 32 + 24 + 8 + 1 = 81.
4. Expand (x − 3)³.
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Coefficients 1, 3, 3, 1 and signed powers of −3: 1, −3, 9, −27. Terms: x³; 3 × x² × (−3) = −9x²; 3 × x × 9 = 27x; −27.
x³ − 9x² + 27x − 27
Check at x = 1: (−2)³ = −8, and 1 − 9 + 27 − 27 = −8.
5. Find the coefficient of x³ in the expansion of (1 + 2x)⁷.
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General term: 7Cr × (2x)ʳ. For x³, r = 3. So 7C3 × 2³ = 35 × 8 = 280.
6. Find the coefficient of x³ in the expansion of (2 − x)⁵.
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General term: 5Cr × 2⁵⁻ʳ × (−x)ʳ. For x³, r = 3. So 5C3 × 2² × (−1)³ = 10 × 4 × (−1) = −40.
7. Find the first three terms, in ascending powers of x, of (1 + 3x)⁸. Use them to estimate 1.003⁸ to 3 decimal places.
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Terms: 1; 8C1 × 3x = 24x; 8C2 × (3x)² = 28 × 9x² = 252x². So the first three terms are 1 + 24x + 252x².
Put x = 0.001: 1 + 0.024 + 0.000252 = 1.024252. To 3 decimal places, 1.003⁸ ≈ 1.024. The terms left out are very small, so this is a sound estimate.
8. The coefficient of x² in (1 + kx)⁶ is 135, and k > 0. Find k.
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6C2 × k² = 15k² = 135, so k² = 9. As k > 0, k = 3.
9. Find the term independent of x in the expansion of (x + 2/x)⁶.
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General term: 6Cr × x⁶⁻ʳ × (2/x)ʳ = 6Cr × 2ʳ × x⁶⁻²ʳ. The power of x is 0 when 6 − 2r = 0, so r = 3. Then 6C3 × 2³ = 20 × 8 = 160.
10. Find the coefficient of x³ in the expansion of (2x − 3)⁵.
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General term: 5Cr × (2x)⁵⁻ʳ × (−3)ʳ. For x³, 5 − r = 3, so r = 2. Then 5C2 × (2x)³ × (−3)² = 10 × 8x³ × 9 = 720x³. The coefficient is 720. The sign is positive because r is even.
11. In the expansion of (1 + ax)⁷, the coefficients of x² and x³ are equal and a ≠ 0. Find a.
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Coefficient of x²: 7C2 a² = 21a². Coefficient of x³: 7C3 a³ = 35a³. Setting them equal: 35a³ − 21a² = 0, so 7a²(5a − 3) = 0. As a ≠ 0, a = 3/5.
Check: 21 × 9/25 = 189/25, and 35 × 27/125 = 945/125 = 189/25.
12. A student writes (x − 2)⁴ = x⁴ − 8x³ + 24x² − 32x − 16. Use x = 0 and x = 1 to show it is wrong, then correct it.
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At x = 0 the bracket gives (−2)⁴ = 16, but the student’s constant term is −16. At x = 1 the bracket gives (−1)⁴ = 1, but the student’s terms give 1 − 8 + 24 − 32 − 16 = −31. Both checks fail.
The last term should be (−2)⁴ = +16, since an even power of a negative number is positive. The corrected expansion is x⁴ − 8x³ + 24x² − 32x + 16. At x = 1: 1 − 8 + 24 − 32 + 16 = 1, which matches.
If you got these wrong
- Wrong coefficient of a power of x, such as 2x³ for 8x³ (questions 3, 5, 10): revisit expanding a positive integer power.
- Wrong term picked, or wrong r (questions 5, 9, 10): revisit finding a specified term.
- Could not form the equation for k or a (questions 8, 11): revisit comparing coefficients to determine a constant.
- Sign errors (questions 4, 6, 10, 12): revisit tracking signs in an alternating expansion.
- No check, or a check that failed and you did not know why (questions 1, 3, 12): revisit checking an expansion using a simple substitution.
Record each slip in the mistake log and retest queue and retry a similar question a few days later. For a wider view of the topic, return to the module overview.
If the same slip returns even after revision, our teachers can look at it with you in online one-to-one Additional Mathematics tuition.