Summation. What does is evaluate to? What is $\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}$ if $a_{n+2}=a_{n+1}+a_{n}$ and $a_{1}=a_{2}=1$?

I found a nice answer, you guys might be curious to see it as well:
Let S be the sum. 
Now $S=\frac{1}{16}+\frac{1}{64}+\frac{2}{256}+\frac{3}{1024}+\frac{5}{4096}$...
Multiply S by 4 to get that 
$4S = \frac{1}{4}+\frac{1}{16}+\frac{2}{64}+\frac{3}{256}+\frac{5}{1024}$...
$3S = \frac{1}{4}+\frac{1}{64}+\frac{1}{256}+\frac{2}{1024}+\frac{3}{4096}$...
$3S = \frac{1}{4}+\frac{1}{4}S$
The rest is simplification... $S=\frac{1}{11}$.
Thanks for all the help guys!
