← Growth and Decay in Sequences › Recurrence Relations › Solutions

Solutions — Recurrence Relations

1. Fluency First 5 terms: t_n+1 = t_n + 6, t₁ = 3. ▶ Show Answer
   t₁ = 3
   t₂ = 3 + 6 = 9
   t₃ = 9 + 6 = 15
   t₄ = 15 + 6 = 21
   t₅ = 21 + 6 = 27
   Sequence: 3, 9, 15, 21, 27 (arithmetic, d = 6)
2. Fluency First 5 terms: t_n+1 = 2t_n, t₁ = 5. ▶ Show Answer
   t₁ = 5
   t₂ = 2 × 5 = 10
   t₃ = 2 × 10 = 20
   t₄ = 2 × 20 = 40
   t₅ = 2 × 40 = 80
   Sequence: 5, 10, 20, 40, 80 (geometric, r = 2)
3. Fluency t_n+1 = 1.1t_n, t₁ = 1000. Find t₄. ▶ Show Answer
   t₂ = 1.1 × 1000 = 1100
   t₃ = 1.1 × 1100 = 1210
   t₄ = 1.1 × 1210 = 1331
4. Fluency Identify arithmetic or geometric. ▶ Show Answer
   (a) t_n+1 = t_n − 5: adds a constant → Arithmetic, d = −5
   (b) t_n+1 = 0.5t_n: multiplies by a constant → Geometric, r = 0.5
5. Understanding t_n+1 = 1.04t_n − 500, t₁ = 10 000. Calculate t₂ through t₅. ▶ Show Answer
   t₂ = 1.04 × 10 000 − 500 = 10 400 − 500 = 9 900
   t₃ = 1.04 × 9 900 − 500 = 10 296 − 500 = 9 796
   t₄ = 1.04 × 9 796 − 500 = 10 187.84 − 500 = 9 687.84
   t₅ = 1.04 × 9 687.84 − 500 = 10 075.35 − 500 = 9 575.35
   
   The loan balance is decreasing each period, meaning the payments are reducing the principal.
6. Understanding Write the recurrence relation for a savings account: starts $2 000, earns 3% interest, $300 deposited each year. ▶ Show Answer
   Interest factor: r = 1.03
   Annual deposit: b = 300
   t_n+1 = 1.03 × t_n + 300,   t₁ = 2000
   
   (Interpretation: at the end of each year, apply 3% interest to current balance, then add the $300 deposit.)
7. Understanding t_n+1 = 0.8t_n + 40, t₁ = 200. Find t₂–t₅ and the long-term value. ▶ Show Answer
   t₂ = 0.8(200) + 40 = 160 + 40 = 200
   Wait — t₂ = t₁ = 200! Let’s check:
   L = 0.8L + 40 → 0.2L = 40 → L = 200
   
   The sequence is already at the fixed point! Every term equals 200.
   
   Interpretation: each dose adds 40 mg, but 20% of the remaining drug is eliminated. The level stabilises at 200 mg because the input and elimination balance out.
8. Understanding t_n+1 = 3t_n − 2, t₁ = 1. Find first 4 terms. ▶ Show Answer
   t₁ = 1
   t₂ = 3(1) − 2 = 1
   t₃ = 3(1) − 2 = 1
   t₄ = 1
   Sequence: 1, 1, 1, 1 — constant!
   
   t₁ = 1 happens to be the fixed point: L = 3L − 2 → −2L = −2 → L = 1. If t₁ ≠ 1, the sequence would grow (since r = 3 > 1).
9. Problem Solving $20 000 loan at 6% annual interest, $2 500 repaid each year. ▶ Show Answer
   Recurrence relation: t_n+1 = 1.06 × t_n − 2500, t₁ = 20 000
   
   t₂ = 1.06 × 20 000 − 2500 = 21 200 − 2500 = $18 700
   t₃ = 1.06 × 18 700 − 2500 = 19 822 − 2500 = $17 322
   t₄ = 1.06 × 17 322 − 2500 = 18 361.32 − 2500 = $15 861
   t₅ = 1.06 × 15 861 − 2500 = 16 812.66 − 2500 = $14 313
   
   Fixed point: L = 1.06L − 2500 → −0.06L = −2500 → L = $41 667
   Since t₁ = $20 000 < L, the balance is decreasing → the loan will eventually be paid off. (It will take approximately 14 years.)
10. Problem Solving t_n+1 = rt_n + b, r = 0.6, b = 80, t₁ = 100. ▶ Show Answer
    (a) If the sequence approaches a fixed value L, then eventually t_n+1 ≈ t_n ≈ L. Substituting: L = rL + b. This is the equation the fixed point must satisfy.
    
    (b) L = 0.6L + 80 → 0.4L = 80 → L = 200
    
    (c) Start near L, say t₁ = 200:
    t₂ = 0.6(200) + 80 = 120 + 80 = 200 ✓ (stays at 200)
    
    Starting at t₁ = 100:
    t₂ = 0.6(100) + 80 = 140
    t₃ = 0.6(140) + 80 = 164
    t₄ = 0.6(164) + 80 = 178.4
    t₅ = 0.6(178.4) + 80 = 187.04
    The sequence is converging toward L = 200, confirming the fixed point.

← Back to Questions