difference of recursive equations Lets have two recursive equations:
\begin{align}
f(0) &= 2  \\
f(n+1) &= 3 \cdot f(n) +  8 \cdot n \\ \\
g(0) &= -2 \\
g(n+1) &= 3 \cdot  g(n) + 12
\end{align}
We want a explicit equation for f(x) - g (x).
I firstly tried to do in manually for first $n$ numbers
\begin{array}{|c|c|c|c|}
\hline
n & f(n) & g(n) & f(n) - g(n) \\ \hline
0 & 2 & -2 &    4 \\ \hline
1 & 6  & 6 &    0 \\ \hline
2 & 26  & 30 &  -4 \\ \hline
3 & 94 & 102 &  -8 \\ \hline
4 & 306  & 318 & -12 \\ \hline
\end{array}
We can deduce that $f(n) - g(n) = 4 - 4n$
But now we have to prove it.
Lets extend recursive equation $f(n)$:
$$f(n) = 3^n \cdot f(0) + 8 \cdot (3^{n+1} \cdot 0 + \dots + 3^{0}(n-1))$$
for $g$ we get $$g(n) = 3^n \cdot g(0) + 12 \cdot (3^{n-1} + \dots + 3^0)$$
We can simply check this by induction but I will skip it, so the question won't be so long.
Now lets  put it together:
$$
f(n) - g(n) = 2 \cdot 3^n + 8 * 3^{n-1} \cdot 0 + ... + 3^0 \cdot (n-1) + 2 \cdot 3^n - 4 \cdot 3 \cdot ( 3^{n-1}+ ... + 3^0)= \\
= 4 \cdot 3^n  + 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^n + 3^{n-1} + ... + 3^1) = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4 \cdot 3^0 = \\
= 8 \cdot (3^{n-1} \cdot 0 + ... + 3^0(n-1)) - 4 \cdot (3^{n-1} + ... + 3^1 + 3^0) + 4
$$
As we can see, we already got the $4$, so to get $-4n + 4$, the rest of the equation must equal $-4n$. But this is where I don't know how to continue.
How to prove that:
$$8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = -4n$$
All I could do is this:
\begin{align}
&8 \cdot (3^{n-1} \cdot 0 + \dots + 3^0(n-1)) - 4 \cdot (3^{n-1} + \dots + 3^1 + 3^0) = \\
&= 4 \cdot (\frac{0}{2}3^{n-1} + \dots + \frac{1}{2} \cdot (n-1) - 4 * (\frac{2}{2}3^{n-1} + \dots + \frac{2}{2}3^0) = \\
&= 4 \cdot (3^{n-1} \cdot (\frac{0- 2}{2}) + \dots + 3^0 \cdot \frac{(n-1)-1}{2}) = \\
&= 4 \cdot (-\frac{2}{2}3^{n-1} + \dots + \frac{n-2}{2}) 
\end{align}
And I made sum function out of it:
$\sum^{n-1}_{i=0}{\frac{i - 2}{2}\cdot 3^{n-1-i}}$
What to do next? Did I go the wrong direction anywhere?
Thank you for your fast responses.

This is an elegant problem. You can use induction on the difference and skip the brute force computation.
Proof by induction:
Let S(n) be the claim that $f(n)-g(n)=4-4n$.
Base case: trivial.
Inductive case: Suppose S(n-1) is true. That is, suppose $f(n-1)-g(n-1)=4-4(n-1)$. Then $f(n)-g(n)=[3f(n-1)+8(n-1)]-[3g(n-1)+12]=3[f(n-1)-g(n-1)]+8n-20.  
$
By induction, that equals:$3[4-4(n-1)]+8n-20=12-12(n-1)+8n-20=12-12n+12+8n-20=24-4n-20=4-4n$
and you're done.
