Solving a recurrence equation with constant term so I have this recurrence equation:
$$G(n) = 2G(n-1) + 100 $$
I believe this a linear non-homogeneous equation of the form $G(n) = LH + F(n) $ where LH is the associated linear homogeneous equation which is G(n-1).
If $F(n)$ is $2n$ we can take a trial solution as $Cn+D$ and solve or if F(n) is say $7^n$ we can take $C\cdot7^n$ and try to solve for constants $C$ and $D$. What do I do if $F(n)$ is a constant like $100$. The initial condition is G($0$) = $2$. 
How do I go about solving this? 
Thanks in advance!

Repeatedly apply the rule $G(n) = 2G(n - 1) + 100$:
\begin{align}
G(n) &= 2G(n - 1) + 1 \cdot 100 \\
&= 2\big( 2G(n - 2) + 100 \big) + 1 \cdot 100 \\
&= 4G(n - 2) + (2 + 1) \cdot 100 \\
&= 4\big( 2G(n - 3) + 100 \big) + (2 + 1) \cdot 100 \\
&= 8G(n - 3) + (4 + 2 + 1) \cdot 100
\end{align}
You should see the following pattern (try proving it by induction):
$$ G(n) = 2^{\color{blue}k}G(n - \color{blue}k) + (2^{\color{blue}k - 1} + \dots + 2^1 + 2^0) \cdot 100 = 2^kG(n-k)+ (2^k - 1)100. $$
Thus
$$ G(n) = 2^{\color{blue}n}G(n - \color{blue}n) + (2^{\color{blue}n} - 1) \cdot 100 = 2^nG(0) + (2^n - 1) \cdot 100. $$
