Suppose P1 = 1, P2 = 2 and for n > 2, Pn = P1 + P2 + … + Pn−1 + 1. Find the values of P1, P2, …, P8. Can you find a simpler recursive formula for Pn? Can you give an explicit formula?
Answer: 1, 2, 4, 8, 16, 32, 64, 128. Simpler rule: Pn = 2Pn−1 (for n ≥ 2, with P1 = 1). Explicit: Pn = 2n−1.
Step-by-step solution
Idea: Work out the first few values to see the pattern (doubling). Then compare the rule for Pn with the rule for Pn−1: they share almost all their terms.
- P3 = 1 + 2 + 1 = 4; P4 = 1 + 2 + 4 + 1 = 8; P5 = 1 + 2 + 4 + 8 + 1 = 16.1 mark
- Continuing: P6 = 32, P7 = 64, P8 = 128. Values: 1, 2, 4, 8, 16, 32, 64, 128.1 mark
- For n ≥ 4: Pn = (P1 + … + Pn−2 + 1) + Pn−1. The bracket is exactly Pn−1, so Pn = Pn−1 + Pn−1 = 2Pn−1. It also holds for n = 2 and 3 (2 = 2 × 1, 4 = 2 × 2).1 mark
- So P1 = 1 and each term doubles: explicit formula Pn = 2n−1.1 mark
P₁ to P₈: 1, 2, 4, 8, 16, 32, 64, 128. Simpler recursive formula: P₁ = 1, Pₙ = 2Pₙ₋₁ for n ≥ 2. Explicit formula: Pₙ = 2ⁿ⁻¹.
Check: P₈ = 1 + 2 + 4 + 8 + 16 + 32 + 64 + 1 = 128 = 2⁷ ✓.
Answer to write in the exam
P3 = 1 + 2 + 1 = 4, P4 = 1 + 2 + 4 + 1 = 8, P5 = 16
P6 = 32, P7 = 64, P8 = 128
∴ 1, 2, 4, 8, 16, 32, 64, 128
Pn = (P1 + … + Pn−2 + 1) + Pn−1 = Pn−1 + Pn−1
∴ Simpler recursive formula: P1 = 1, Pn = 2Pn−1 for n ≥ 2
∴ Explicit formula: Pn = 2n−1
Common mistakes that cost marks
- Forgetting the + 1 at the end of the rule (giving P3 = 3).
- Writing the explicit formula as 2n, which gives P1 = 2. Check with n = 1: 20 = 1.
- Stopping at ‘the terms double’ without showing why. Show that the sum part of Pn equals Pn−1.
How this can come in the exam
MCQ (1 mark)
With Pn = 2n−1, the value of P11 is
- 1024
- 2048
- 512
- 22
Show answer
(A) 1024
210 = 1024.
Try one yourself
Suppose Q1 = 2 and for n ≥ 2, Qn = Q1 + Q2 + … + Qn−1 + 2. Find Q1 to Q5 and an explicit formula.
Show answer
2, 4, 8, 16, 32: each term doubles, so Qn = 2n.
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