1. ## reccurence relations

In my notes I have the expression:
$a_n = 3a_{n-1} + 2a_{n-2}$

with
$a_0 = 1, a_1 = 2$

but I can't see how we get $a_0 = 1$

When I try I get
$3a_{0-1} + 2a_{0-2}
= 3*-1 + 2*-2
= -3 + -4
= -7
$

so why is
$+ 2a_{n-2}
= 4$

to get -3 + 4 = 1

obviously I am not understanding something...
can you point out where I am going wrong please?
Thanks

2. Go to this web link. You can change values for $f(1)$ click the equals sign. You will see how the recursion changes.

Now why? Well we need to know how the initial values because we have two definiting terms.

3. That's a great link, I'm sorry but I still don't get it.
if I'm looking for f(4)
then
f(2) = 3a(2-1) + 2a(2-2)
f(2) = 3*a(1) + 2*a(0) <---- where a(1) = 2 and a(0) = 1
f(2) = 3*2 + 2*1
f(2) = 8

f(3) = 3a(3-1) + 2a(3-2)
f(3) = 3*a(2) + 2*a(1) <----a(2) = 8, a(1) =2
f(3) = 3*8 + 2*2
f(3) = 28

f(4) = 3a(4-1) + 2a(4-2)
f(4) = 3*a(3) + 2*a(2) <-----a(3)=28, a(2) = 8
f(4) = 3*28 + 2*8
f(4) = 100

is that right?
(I think I was reading too deep into things when i was trying to figure out how a(0) = 1, I guess it's just a given because a(0) =1 and a(1) = 2 are initial values, we don't need to work out how to get them, we just use them to find higher values??)

Just wandering if I have worked through this correctly or not?