tₙ = 7tₙ₋₁ – 10tₙ₋₂
t₀=5 t₁=16
Recurrence relation
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Associate homogeneous solution is ar - 5ar-1 + 6ar-2
The characteristic equation of its associated homogeneous relation is
x2 - 5 x + 6 = 0
(x - 3) (x - 2) = 0
∴ x = 3 or x = 2
The solution of associated homogeneous recurrence relationan = 6an-2 - an-1 is
Ah = A1(2)r + A2(3)r
f(r) = 1×4r it is of the form and 4 is not a root.Therefore it's particular solution is A4r
General solution of recurrence relation is A1(2)r + A2(3)r + A4r
After substitution partial solution in recurrence relation
(A4r) - 5(A4r-1) + 6(A4r-2) = 4r
A - (5A/4) + (6A/16) = 1
∴ A = 8
Therefore general solution is A1(2)r + A2(3)r + 8.4r
I hope it will help you.
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