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Solve aₙ=2aₙ₋₁+2n²

Short answer

Homogeneous: α·2^n. Particular: -2n^2 - 8n - 12. General solution: a_n = α·2^n - 2n^2 - 8n - 12. With a1=4, α=13, so a_n = 13·2^n - 2n^2 - 8n - 12.

Find all solutions of the recurrence relation aₙ = 2aₙ₋₁ + 2n² with initial condition a₁ = 4.

The relation aₙ = 2aₙ₋₁ + 2n² is linear, non-homogeneous, of order 1.

Step 1. Homogeneous part

aₙ = 2aₙ₋₁ gives r = 2, so aₙ⁽ʰ⁾ = α·2ⁿ.

Step 2. Particular solution

F(n) = 2n² is a polynomial of degree 2 and 1 is not a root, so try aₙ⁽ᵖ⁾ = pn² + qn + r.

pn² + qn + r = 2[p(n-1)² + q(n-1) + r] + 2n²
             = 2pn² - 4pn + 2p + 2qn - 2q + 2r + 2n²
n²:     p = 2p + 2            ->  p = -2
n:      q = -4p + 2q          ->  q = 4p = -8
const:  r = 2p - 2q + 2r      ->  r = -2p + 2q = 4 - 16 = -12

aₙ⁽ᵖ⁾ = −2n² − 8n − 12.

Step 3. General solution (all solutions)

aₙ = α·2ⁿ − 2n² − 8n − 12, where α is any constant.

Step 4. Initial condition a₁ = 4

2α − 2 − 8 − 12 = 4, so 2α = 26 and α = 13.

Answer: aₙ = 13·2ⁿ − 2n² − 8n − 12

Check: a₁ = 26 − 2 − 8 − 12 = 4; a₂ = 52 − 8 − 16 − 12 = 16 = 2(4) + 2(4) ✓; a₃ = 104 − 18 − 24 − 12 = 50 = 2(16) + 18 ✓.

✓ Answer checked by Claude Opus 5.5 (high effort)

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