How to calculate the explicit form of a recursive function?

Question

I have this recursive function:

f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
f(1) = 2
f(2) = 8

I know from experience that explicit form of it would be:

f(n) = 3 ^ n - 1  // pow(3, n) - 1

I wanna know if there's any way to prove that. I googled a bit, yet didn't find anything simple to understand. I already know that generation functions probably solve it, they're too complex, I'd rather not get into them. I'm looking for a simpler way.

P.S. If it helps I remember something like this solved it:

f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
// consider f(n) = x ^ n
x ^ n = 2 * x ^ (n-1) + 3 * x ^ (n-2) + 4

And then you somehow computed x that lead to explicit form of the recursive formula, yet I can't quite remember

Answer 1

f(n) = 2 * f(n-1) + 3 * f(n-2) + 4
f(n+1) = 2 * f(n) + 3 * f(n-1) + 4

f(n+1)-f(n) = 2 * f(n) - 2 * f(n-1) + 3 * f(n-1) - 3 * f(n-2)
f(n+1) = 3 * f(n) + f(n-1) - 3 * f(n-2)

Now the 4 is gone. As you said the next step is letting f(n) = x ^ n

x^(n+1) = 3 * x^n + x^(n-1) - 3 * x^(n-2)

divide by x^(n-2)

x^3 = 3 * x^2 + x - 3
x^3 - 3 * x^2 - x + 3 = 0

factorise to find x

(x-3)(x-1)(x+1) = 0
x = -1 or 1 or 3

f(n) = A * (-1)^n + B * 1^n + C * 3^n
f(n) = A * (-1)^n + B + C * 3^n

Now find A,B and C using the values you have

f(1) = 2; f(2) = 8; f(3) = 26

f(1) = 2 = -A + B + 3C
f(2) = 8 = A + B + 9C
f(3) = 26 = -A + B + 27C

solving for A,B and C:

f(3)-f(1) = 24 = 24C      => C = 1
f(2)-f(1) = 6 = 2A + 6    => A = 0
2 = B + 3                 => B = -1

Finally

f(n) = 3^n - 1