# Solving a Linear Homogeneous Recurrence Relation

1. Dec 12, 2011

### pavel329

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

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

Last edited by a moderator: Dec 12, 2011
2. Dec 12, 2011

### Office_Shredder

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

Last edited by a moderator: Dec 12, 2011
3. Dec 12, 2011

### pavel329

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.

Last edited by a moderator: Dec 12, 2011
4. Dec 12, 2011

### Dick

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

Last edited by a moderator: Dec 12, 2011
5. Dec 12, 2011

### pavel329

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?

Last edited by a moderator: Dec 12, 2011
6. Dec 12, 2011

### Dick

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.

7. Dec 12, 2011

### pavel329

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

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

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

8. Dec 12, 2011

### Dick

It sure is.

9. Dec 12, 2011

### pavel329

Thank you very much for your help.