The Binomial Theorem — Solutions

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1. Expand using the binomial theorem. Fluency

   1. (a) (x + 1)³ = x³ + 3x² + 3x + 1
   2. (b) (x + y)⁴ = x&sup4; + 4x³y + 6x²y² + 4xy³ + y&sup4;
   3. (c) (a − b)³ Substitute y = −b: a³ − 3a²b + 3ab² − b³
   4. (d) (1 + x)⁵ = 1 + 5x + 10x² + 10x³ + 5x&sup4; + x&sup5;
   5. (e) (x − 2)⁴ = x&sup4; − 4(2)x³ + 6(4)x² − 4(8)x + 16 = x&sup4; − 8x³ + 24x² − 32x + 16

2. Expand and simplify. Fluency

   1. (a) (2x + 1)³ = (2x)³ + 3(2x)²(1) + 3(2x)(1)² + 1 = 8x³ + 12x² + 6x + 1
   2. (b) (x + 3)⁴ = x&sup4; + 4(3)x³ + 6(9)x² + 4(27)x + 81 = x&sup4; + 12x³ + 54x² + 108x + 81
   3. (c) (2 − x)⁴ = 16 − 4(8)x + 6(4)x² − 4(2)x³ + x&sup4; = 16 − 32x + 24x² − 8x³ + x&sup4;
   4. (d) (3x + 2)³ = 27x³ + 3(9x²)(2) + 3(3x)(4) + 8 = 27x³ + 54x² + 36x + 8

3. General term and specific terms. Fluency

   1. (a) T₄ of (x + 2)⁶: r = 3. T₄ = ⁶C₃ x³ (2)³ = 20 × 8x³ = 160x³
   2. (b) T₅ of (1 + x)⁷: r = 4. T₅ = ⁷C₄ x&sup4; = 35x&sup4;
   3. (c) T₃ of (x − 1)⁸: r = 2. T₃ = ⁸C₂ x&sup6;(−1)² = 28x&sup6;
   4. (d) T₂ of (2x + 3)⁵: r = 1. T₂ = ⁵C₁(2x)&sup4;(3) = 5 × 16x&sup4; × 3 = 240x&sup4;

4. Specific coefficients. Fluency

   1. (a) Coefficient of x³ in (1 + x)⁷. T_r+1 = ⁷C_rx^r. For x³: r = 3. Coefficient = ⁷C₃ = 35.
   2. (b) Coefficient of x² in (x + 3)⁵. T_r+1 = ⁵C_rx^5−r(3)^r. For x²: 5−r = 2, r = 3. Coefficient = ⁵C₃(3)³ = 10 × 27 = 270.
   3. (c) Coefficient of x&sup4; in (2x − 1)⁶. T_r+1 = ⁶C_r(2x)^6−r(−1)^r. For x&sup4;: 6−r = 4, r = 2. Coefficient = ⁶C₂(2)⁴(−1)² = 15 × 16 = 240.
   4. (d) Constant term in (x + 2/x)⁶. T_r+1 = ⁶C_r x^6−r(2/x)^r = ⁶C_r 2^r x^6−2r. For constant: 6−2r = 0, r = 3. Constant = ⁶C₃ × 2³ = 20 × 8 = 160.

5. Expand and collect. Fluency

   1. (a) (1 + x)⁴ + (1 − x)⁴ (1+x)&sup4; = 1 + 4x + 6x² + 4x³ + x&sup4;
      (1−x)&sup4; = 1 − 4x + 6x² − 4x³ + x&sup4;
      Sum = 2 + 12x² + 2x&sup4; = 2(1 + 6x² + x&sup4;)
   2. (b) (1 + x)⁴ − (1 − x)⁴ Difference = 8x + 8x³ = 8x(1 + x²)

6. Applying the theorem. Understanding

   1. (a) 1.1⁴ = (1 + 0.1)⁴. = 1 + 4(0.1) + 6(0.01) + 4(0.001) + 0.0001 = 1 + 0.4 + 0.06 + 0.004 + 0.0001 = 1.4641
   2. (b) 0.99⁵ = (1 − 0.01)⁵. = 1 − 5(0.01) + 10(0.0001) − 10(0.000001) + 5(0.00000001) − 0.0000000001
      ≈ 1 − 0.05 + 0.001 − 0.00001 + 0.0000000 ≈ 0.95099
   3. (c) (1 + √2)⁴ + (1 − √2)⁴. Using part (a) result pattern with x = √2:
      = 2(1 + 6(√2)² + (√2)&sup4;) = 2(1 + 6(2) + 4) = 2(1 + 12 + 4) = 2(17) = 34

7. Expansion of (1+x)⁶ and largest terms. Understanding

   1. (a) (1 + x)⁶ and greatest coefficient. = 1 + 6x + 15x² + 20x³ + 15x&sup4; + 6x&sup5; + x&sup6;.
      Greatest coefficient is 20 (for x³, the middle term).
   2. (b) T₃ = T₄ in (2 + x)⁵. T₃ = ⁵C₂(2)³x² = 40x².
      T₄ = ⁵C₃(2)²x³ = 40x³.
      40x² = 40x³ → x = 1.

8. Connection to identities. Understanding

   1. (a) Row sum = 2ⁿ. Set x = 1, y = 1 in (x+y)ⁿ: 2ⁿ = ⁿC₀ + ⁿC₁ + … + ⁿC_n ✓
   2. (b) Alternating sum = 0. Set x = 1, y = −1: (1 − 1)ⁿ = 0ⁿ = 0 = ⁿC₀ − ⁿC₁ + … ✓ (for n ≥ 1)

9. Finding unknown coefficients. Problem Solving

   1. (a) Coefficient of x³ in (1 + kx)⁶ = 160. T₄: ⁶C₃(kx)³ = 20k³x³. So 20k³ = 160 → k³ = 8 → k = 2.
   2. (b) Coefficient of x² in (2 + ax)⁴ = 96. T₃: ⁴C₂(2)²(ax)² = 6 × 4 × a²x² = 24a²x². So 24a² = 96 → a² = 4 → a = ±2.
   3. (c) Coefficient of x is 12, coefficient of x² is 60. T₂ = np(x) → np = 12. T₃ = ⁿC₂p²x² = n(n−1)p²/2 = 60.
      From np = 12: p = 12/n. Substitute: n(n−1)(144/n²)/2 = 60 → 144(n−1)/(2n) = 60 → 72(n−1)/n = 60 → 72n−72 = 60n → 12n = 72 → n = 6, p = 2.

10. Combining expansions. Problem Solving

    1. (a) Coefficient of x&sup5; in (1+x)⁴(1+x)³. (1+x)&sup4;(1+x)³ = (1+x)⁷. Coefficient of x&sup5; = ⁷C₅ = 21.
    2. (b) Coefficient of x³ in (1 + x + x²)(1 + x)⁴. (1+x)&sup4; = 1 + 4x + 6x² + 4x³ + …
       Coefficient of x³ in product = 1(4) + 1(6) + 1(4) = 4 + 6 + 4 = 14. (from 1·x³ + x·x² + x²·x terms)
    3. (c) Middle term of (1+x)²ⁿ. (1+x)²ⁿ has 2n+1 terms. Middle term is the (n+1)th: T_n+1 = ²ⁿC_n(1)ⁿxⁿ = ²ⁿC_nxⁿ. Coefficient = ²ⁿC_n ✓