# Recurrence relations discrete math problem

1. May 16, 2012

### charmedbeauty

1. The problem statement, all variables and given/known data

Find the general solution to the following recurrence relations (defined n≥2).

c) an=6an-1-9an-2+8n+4

2. Relevant equations

3. The attempt at a solution

an=6an-1-9an-2+8n+4

8n+4= an -6an-1+9an-2

R2-6R+9=0

R=3,3

So hn=A(3)n+B(3)n

Assume pn=Cn+Cn2 → This is where I got stuck!

What should my assumption be??

I thought this because RHS =8n+4

so for the 4 I used Cn since 4 is a constant so I put an extra n term...likewise for 8n.

but it turned out to be a mess!

HELP

2. May 16, 2012

### tiny-tim

hi charmedbeauty!
no, if a recurrence relation has a root R repeated k times, the k independent solutions are niRn (0≤i<k)

3. May 16, 2012

### charmedbeauty

hmm im a little confused what should i be?

so i replace hn=A(3)n+B(3)n

with hn=A(3n)+Bn(3n)??

4. May 17, 2012

### tiny-tim

yup!

(and you can leave out those brackets … A3n + Bn3n )

5. May 17, 2012

Thanks.