Solve the recurrence relation a_n = -2a_n-1 + 15a_n-2, n ≥ 2, given a₀ = 1, a₁ = -1.
x² + 2x - 15, the distinct roots 3 and -5, so a_n = C₁(3^n) + C₂(-5)^n. The initial condition gives a₀ = 1 = C₁ - C₂, a₁ = -1 = 3C₁ - 5C₂. We obtain C₁ = C₂ = 1/2 and so a_n = 1/2(3^n) + 1/2(-5)^n.
My question is how does C₁ = C₂ = 1/2 can some please how do you derive to this answer because I'm confused.Thank you for any help.
There wasa typo in the equations derived from the initial conditions. You should have:
The initial condition gives
a₀ = 1 = C₁ + C₂,
a₁ = -1 = 3C₁ - 5C₂
It then easily follows that
C₁ = C₂ = 1/2
Sorry I still don't get it. Can someone please explain?
To find the values of C₁ and C₂, we can use the initial conditions a₀ = 1 and a₁ = -1.
Substitute n = 0 into the expression for a_n:
a₀ = 1/2(3^0) + 1/2(-5)^0
= 1/2 + 1/2
= 1.
Since we are given that a₀ = 1, we now have the equation C₁ - C₂ = 1.
Substitute n = 1 into the expression for a_n:
a₁ = 1/2(3^1) + 1/2(-5)^1
= 3/2 - 5/2
= -1.
Since we are given that a₁ = -1, we now have the equation 3C₁ - 5C₂ = -1.
We now have a system of two equations with two unknowns:
C₁ - C₂ = 1
3C₁ - 5C₂ = -1.
One way to solve this system of equations is to multiply the first equation by 3 and the second equation by 1, so that the coefficients of C₁ will be the same in both equations:
3(C₁ - C₂) = 3(1)
3C₁ - 3C₂ = 3.
Now we can subtract the second equation from the first equation:
(3C₁ - 3C₂) - (3C₁ - 5C₂) = 3 - (-1)
3C₁ - 3C₂ - 3C₁ + 5C₂ = 4
2C₂ = 4
C₂ = 2.
Substitute this value of C₂ into the first equation:
C₁ - 2 = 1
C₁ = 3.
So, we have found that C₁ = 3 and C₂ = 2.
Therefore, the solution to the recurrence relation is:
a_n = 3(3^n)/2 + 2(-5)^n/2.