SAT/ GCSE-Level Recurrence Relation Problem

[odolwa99]

Homework Statement

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 textbook 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.

Homework Equations

Q. Find the first five terms of the sequence:

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

The Attempt at a Solution

Attempt:

Solve u_n+1 where u_n = 3u_n-1 - u_n-2
=> 3u_(n+1)-1 - u_(n+1)-2

Begin by substituting 3 (i.e. u₂) for u_n:
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.

 

[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 $u_{4}$
$$u_{4}=2u_{3}+u_{2}=2\times 7+3=17$$

 

[odolwa99]

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

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

 

[hunt_mat]

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

 

[odolwa99]

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

 

[hunt_mat]

it's why I help here.