<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] You have $G_n=\frac{1}{4}-\frac{1}{4}G_{n-1}$. With the ansatz $G_n=r^n$ we have for the homogeneous
$$r^n=-\frac{1}{4}r^{n-1}$$
$$r=-\frac{1}{4}$$
and for the particular
$$K=\frac{1}{4}-\frac{1}{4}K$$
$$K=\frac{1}{5}$$
so the general solution is $G_n=\frac{1}{5}+B\left(-\frac{1}{4}\right)^n$ and we plug in the initial conditions to get $0=\frac{1}{5}+B$, so $B=-\frac{1}{5}$.
Thus $G_n=\frac{1}{5}\left( 1-\left(-\frac{1}{4}\right)^n\right)$.
By the way, the answer you were given is not consistent with the initial values. If you plug in $n=0$ you get $G_0=\frac{3}{20}$.
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