Solve aₙ=5aₙ₋₁−6aₙ₋₂+3n+2ⁿ

Reading of the question: the recurrence is of order 2, so two initial values a₀ = 0, a₁ = 1 determine the solution. The third value a₂ = 2 given in the question is not consistent with the relation (the recurrence gives a₂ = 5(1) − 6(0) + 6 + 4 = 15). So I use a₀ and a₁ and apply the relation for n ≥ 2.

Step 1. Homogeneous part

aₙ = 5aₙ₋₁ − 6aₙ₋₂ gives r² − 5r + 6 = 0, so r = 2, 3. aₙ⁽ʰ⁾ = α·2ⁿ + β·3ⁿ.

Step 2. Particular solution

F(n) = 3n + 2ⁿ. Handle the two terms separately.

(a) For 3n (1 is not a root): try pn + q. Substituting:

pn + q = 5[p(n-1)+q] - 6[p(n-2)+q] + 3n
pn + q = -pn + 7p - q + 3n
n:      p = -p + 3     ->  p = 3/2
const:  q = 7p - q     ->  q = 7p/2 = 21/4

(b) For 2ⁿ: 2 is a simple root of the characteristic equation, so try c·n·2ⁿ.

c n 2ⁿ = 5c(n-1)2ⁿ⁻¹ - 6c(n-2)2ⁿ⁻² + 2ⁿ        (divide by 2ⁿ⁻²)
4cn = 10c(n-1) - 6c(n-2) + 4 = 4cn + 2c + 4
0 = 2c + 4   ->  c = -2

Particular solution: aₙ⁽ᵖ⁾ = (3/2)n + 21/4 − 2n·2ⁿ.

Step 3. General solution

aₙ = α·2ⁿ + β·3ⁿ + (3/2)n + 21/4 − n·2ⁿ⁺¹

Step 4. Use initial conditions

n = 0:  α + β + 21/4 = 0                    ->  α + β = -21/4
n = 1:  2α + 3β + 3/2 + 21/4 - 4 = 1       ->  2α + 3β = -7/4

Subtracting 2×(first) from the second: β = −7/4 + 42/4 = 35/4, then α = −21/4 − 35/4 = −14.

Answer: aₙ = −14·2ⁿ + (35/4)·3ⁿ + (3/2)n + 21/4 − n·2ⁿ⁺¹

Check: this gives a₀ = 0, a₁ = 1, a₂ = 15, a₃ = 86, a₄ = 368, which agrees with the recurrence (verified in Python for n up to 11).