<s>[INST] Solving Recurrence Relation problem I am trying to solve the recurrence relation:
$$G_n = \frac{1-G_{n-1}}{4}$$
$$G_0 = 0$$
$$G_1 = \frac{1}{4}$$
I am told that the answer is
$$G_n = \frac{1}{5}\left(1+\left(\frac{-1}{4}\right)^{n+1}\right)$$
I have found the characteristic equation to be 
$$G_n = B\left(\frac{-1}{4}\right)^{n}$$
But then I am not sure how to continue.
 [/INST] By rewriting the recurrence formula we have $G_n=\dfrac{1}{4}-\dfrac{1}{4}G_{n-1}$.
If we write the first elements of the sequence we get:
$$\begin{align*}
G_0&=0,  G_1=\dfrac{1}{4} \\
G_2&=\dfrac{1}{4}-\dfrac{1}{4}G_1 =\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2\\
G_3&=\dfrac{1}{4}-\dfrac{1}{4}G_2 =\dfrac{1}{4}-\dfrac{1}{4}\left(\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2\right)\\
& \hspace{2.3cm}=\dfrac{1}{4}-\left(\dfrac{1}{4}\right)^2+\left(\dfrac{1}{4}\right)^3\\
\vdots & \hspace{3cm}\vdots\\
G_n&=^{(?)}\sum_{k=1}^n (-1)^{k+1}\dfrac{1}{4^k}=(-1)\cdot \sum_{k=1}^n \left(\dfrac{-1}{4}\right)^k
\end{align*}$$
Once you have proved the recurrence formula then by properties of geometric series you get
$$G_n=(-1)\cdot \sum_{k=1}^n \left(\dfrac{-1}{4}\right)^k=(-1)\left(\dfrac{1-(-\frac{1}{4})^{n+1}}{1-(-\frac{1}{4})}-1\right)=\dfrac{-1}{1+\frac{1}{4}}\left(1-\left(-\frac{1}{4}\right)^{n+1}-\dfrac{5}{4}\right)=\dfrac{-4}{5}\left(-\dfrac{1}{4}-\left(-\frac{1}{4}\right)^{n+1}\right)=\dfrac{1}{5}\left(1-\left(-\frac{1}{4}\right)^{n}\right)$$
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