7. a) Given the recurrence relation a, = 6am-1 11an-2 + 6an-3 and the initialconditions ao = 2, a, = 5 and az = 15, find the solution.
Answers
Given : aₙ = 6aₙ₋₁ - 11aₙ₋₂ + 6aₙ₋₃
a₀ = 2 , a₁ = 5 , a₂ = 15
To Find : Solution
Solution:
aₙ = 6aₙ₋₁ - 11aₙ₋₂ + 6aₙ₋₃
aₙ = 6aₙ₋₁ - 11aₙ₋₂ + 6aₙ₋₃
=> aₙ = α₁r₁ⁿ + α₂r₂ⁿ + α₃r₃ⁿ
where r₁ , r₂ and r₃ are roots of
r³ - 6r² + 11r - 6 = 0
=> (r - 1) ( r² - 5r + 6) = 0
=> (r - 1)(r - 2)(r - 3) = 0
Hence
aₙ = α₁ + α₂2ⁿ + α₃3ⁿ
a₀ = α₁ + α₂ + α₃ = 2 Eq1
a₁ = α₁ + 2α₂ + 3α₃ = 5 Eq2
a₂ = α₁ + 4α₂ + 9α₃ = 15 Eq3
Eq2 - Eq1
=> α₂ + 2α₃ = 3
Eq3 - Eq2
=> 2α₂ + 6α₃ = 10 => α₂ + 3α₃ = 5
=> α₃ = 2
=> α₂ = - 1
Hence α₁ = 1
=> aₙ = 1 - 2ⁿ + 2. 3ⁿ
aₙ = 2. 3ⁿ - 2ⁿ + 1
a₀ = 2 - 1 + 1 = 2 Verified
a₁ = 6 - 2 + 1 = 5 Verified
a₂ = 18 - 4 + 1 = 15 Verified
a₃ = 54 - 8 + 1 = 47
a₃ = 6a₂ - 11a₁ + 6a₀
=> a₃ = 6(15) - 11(5) + 6(2)
=> a₃ = 90 - 55 + 12
=> a₃ = 47 Verified
aₙ = 2. 3ⁿ - 2ⁿ + 1 is the required solution .
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