# Homework Help: SAT/ GCSE-Level Recurrence Relation Problem

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1. Jul 16, 2011

### odolwa99

1. The problem statement, all variables and given/known data
Hi! This is my first time on the site. I look forward to working with everyone…but hopefully not too much, assuming I‘m learning things correctly. :P

My question pertains to Recurrence Relations, so here it goes…

Foreword: The text book I’m using actually supplies the answer to the question, so I already have a point of reference, but my attempt does not match up with the answers. I believe my approach is essentially correct, as it has yielded the correct answers for a similar question beforehand. Answer is: 1, 3, 7, 17, 41

Please note that I am beginning the question from u3, as we already have the values for u1 and u2.

2. Relevant equations

Q. Find the first five terms of the sequence:

u1 = 1, u2 = 3 and un = 2un-1 + un-2

3. The attempt at a solution

Attempt:

Solve un+1 where un = 3un-1 - un-2
=> 3u(n+1)-1 - u(n+1)-2

Begin by substituting 3 (i.e. u2) for un:
If n = 1 then u3 = 2((3+1) - 1) + ((3+1) -2) => 2(4-1) + (4-2) => 6 + 2
Ans.: u3 = 8... but should be 7!!!

Proceeding with u3 as 7, not 8...

If n = 2 then u4 = 2((7+1) -1) + ((7+1) -2) => 2(8-1) + (8-2) => 14 + 6
Ans.: u4 = 20... But should be 17!!!

Note, I am omitting solution of u5 for brevity’s sake.

I‘m sure the answer is staring me in the face, but I just can’t seem to figure it out!
Can anyone help?

Thanks.

Last edited: Jul 16, 2011
2. Jul 16, 2011

### hunt_mat

Not too sure what you're going here but let's calculate $u_{3}$ from the recurrence relation.
$$u_{3}=2u_{2}+u_{1}=2\times 3+1=7$$
Working for $u_{4}$
$$u_{4}=2u_{3}+u_{2}=2\times 7+3=17$$

3. Jul 16, 2011

### odolwa99

Woah, that was easier than I was making it! Thank you.

One final question though, why is the value of u1 subbed into un-2 and u2 into un-1?

4. Jul 16, 2011

### hunt_mat

you're finding n=3, so n-1=2 and n-2=1.

5. Jul 16, 2011

### odolwa99

Thank you very much. You've really helped me out!

6. Jul 16, 2011

### hunt_mat

it's why I help here.

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