Recursie Algorithms. Is my solution ok?

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The discussion focuses on solving the recurrence relation an = 7an-1 - 12an-2 with initial conditions a0 = 3 and a1 = 10. The user expresses uncertainty about deriving a non-recursive formula for an and attempts to calculate a2 but arrives at an incorrect value. Participants clarify that the goal is to find an exact algebraic equation rather than a recursive solution. There is an emphasis on the importance of algebraic accuracy in solving the problem. The user acknowledges the need to improve their algebra skills to correctly derive the solution.
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This is the question I must solve;
Solve the given recurrence relation for the given inital conditions.
(This means give a formula in terms of n, not in terms of previous entries)
a[sub]n[/sub] = 7a[sub]n-1[/sub] - 12a[sub]n-2[/sub]
a[sub]0[/sub] = 3 a[sub]1[/sub] = 10
Now I am not sure what that means but I think this will solve the question 

let me know if I am not even close;
Input = n
Output = X(n)

procdure find(n)
 if n = 3 or n = 10 then
  return (n)
 return(find(n-1)+find(n-2))
end find(n)
 
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I'm fairly certain that what they are looking for is a single equation for a as a function of n. Simple example:

Given: an = an-1 + 1
and a0 = 11

Then:

an = 11+n
 
Code:
Solve the given recurrence relation for the given inital conditions.
(This means give a formula in terms of n, not in terms of previous entries)
a[sub]n[/sub] = 7a[sub]n-1[/sub] - 12a[sub]n-2[/sub]
a[sub]0[/sub] = 3 a[sub]1[/sub] = 10

So;
a[sub]2[/sub] = 7a[sub]2-1[/sub] - 12a[sub]2-2[/sub]
a[sub]2[/sub] = 7a[sub]1[/sub] - 12a[sub]0[/sub]
a[sub]2[/sub] = 7(10) - 12(3)
a[sub]2[/sub] = 70 - 36
a[sub]2[/sub] = 34?
 
Oh not that again I thought I was done with that. Ok I know what to do thanks again.
 
I come up with all mixed fractions when I get a using the same method as I did on that problem? So that does not work.
 
It works. You need to work on your algebra.
 
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