Discrete StructureTU Board 2081
Solve the recurrence relation a n = a n−1 + 2a n−2 with initial conditions a 0 = 2 and a 1 = 7.
5Answer
Recurrence: aₙ = aₙ₋₁ + 2aₙ₋₂, with a₀ = 2 and a₁ = 7.
This is a linear homogeneous recurrence of degree 2 with constant coefficients.
Step 1: characteristic equation
Put aₙ = rⁿ: r² = r + 2, so r² − r − 2 = 0, which factors as (r − 2)(r + 1) = 0.
The roots are r = 2 and r = −1 (distinct).
Step 2: general solution
aₙ = α·2ⁿ + β·(−1)ⁿ
Step 3: use the initial conditions
- n = 0: α + β = 2
- n = 1: 2α − β = 7
Adding the two equations: 3α = 9, so α = 3, and then β = −1.
Solution
aₙ = 3·2ⁿ − (−1)ⁿ
Check: from the recurrence, a₂ = a₁ + 2a₀ = 7 + 4 = 11. From the formula, a₂ = 3·4 − 1 = 11. ✓
Discussion
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