Solving a Linear Homogeneous Recurrence Relation

[pavel329]

1. Solve the following recurrence relation.
a_n - 5a_n-1 + 6a_n-2 = 0, n ≥ 2, a₀ = 1, a₁ = 3







3. My attempt
I changed it to 0 = tⁿ - 5t^n-1 + 6t^n-2
Don't know where to go from there.

 

[Office_Shredder]

Do you see something you can factor out of that polynomial?

 

[pavel329]

Do you see something you can factor out of that polynomial?

I'm not sure how to factor anything with (n-1) or (n-2) as an exponent.

The book turned a similar equation into t²-t-6.
Which made no sense to me.

 

[Dick]

I'm not sure how to factor anything with (n-1) or (n-2) as an exponent.

The book turned a similar equation into t²-t-6.
Which made no sense to me.

First factor out t^(n-2). It divides all three terms, right?

 

[pavel329]

First factor out t^(n-2). It divides all three terms, right?

Ok so I now have it down to
t=2,3.
Now I assume i need to put that into an equation for the original problem.
Which I have no clue how to do.

But where does the n≥2 stand?
Does that affect my final outcome?

 

[Dick]

Ok so I now have it down to
t=2,3.
Now I assume i need to put that into an equation for the original problem.
Which I have no clue how to do.

But where does the n≥2 stand?
Does that affect my final outcome?

Are you sure you've seen examples of this before? They should have given you one before they gave you this problem. The t=2,3 tells you the solution is a_n=A*2^n+B*3^n. You know a0 and a1. Use those to find A and B.

 

[pavel329]

Are you sure you've seen examples of this before? They should have given you one before they gave you this problem. The t=2,3 tells you the solution is a_n=A*2^n+B*3^n. You know a0 and a1. Use those to find A and B.

I've only seen one example.
s_n = s_n-1 + 6s_n-2, s₀ = 4 s₁ = 7

However this equation put a 0 in place of the a_n and added the n≥2.
Which maybe I shouldn't have paid attention to.

But my final answer is: A=1 B=0. So a_n = 3ⁿ. Is this correct?

 

[Dick]

I've only seen one example.
s_n = s_n-1 + 6s_n-2, s₀ = 4 s₁ = 7

However this equation put a 0 in place of the a_n and added the n≥2.
Which maybe I shouldn't have paid attention to.

But my final answer is: A=1 B=0. So a_n = 3ⁿ. Is this correct?

It sure is.

 

[pavel329]

It sure is.

Thank you very much for your help.