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!

Another approach using generating functions:
$\begin{align}
F(x) &= \sum_{n=0}^{\infty} G(n) x^n \\
F(x) &= 2x^0 + \sum_{n=1}^{\infty} (2G(n-1) + 100) x^n \\
F(x) &= 2 + 2\sum_{n=1}^{\infty} G(n-1)x^n + 100\sum_{n=1}^{\infty}  x^n \\
F(x) &= 2 + 2x\sum_{n=1}^{\infty} G(n-1)x^{n-1} + 100(-x^0 + \sum_{n=0}^{\infty}  x^n) \\
F(x) &= 2 + 2x\sum_{n=0}^{\infty} G(n)x^{n} + 100(-1 + \sum_{n=0}^{\infty}  x^n) \\
F(x) &= 2 + 2xF(x) + 100\left(-1 + \frac{1}{1-x}\right) \\
F(x) &= 102 \cdot \frac{1}{1 - 2 x}  - 100 \cdot \frac{1}{1 - x}\\
\end{align}$
Take the $n$th coefficient of this generating function and we get:
$G(n) = 102 \cdot 2^n  - 100$
