# Solving a Linear Homogeneous Recurrence Relation

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.

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

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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.

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Dick
Homework Helper
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?

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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?

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Dick
Homework Helper
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.

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.
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?

Dick
Homework Helper
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?
It sure is.

It sure is.
Thank you very much for your help.