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 $a_{1}=a_{2}=1$?
 A: $a_{n+2}=a_{n+1}+a_{n}$
with
$a_1=a_2 = 1$.
Let
$f(x)
=\sum_{n=1}^{\infty} a_nx^n
$.
$xf(x)
=\sum_{n=1}^{\infty} a_nx^{n+1}
=\sum_{n=2}^{\infty} a_{n-1}x^{n}
$
and
$x^2f(x)
=\sum_{n=1}^{\infty} a_nx^{n+2}
=\sum_{n=3}^{\infty} a_{n-2}x^{n}
$
so
$\begin{array}\\
xf(x)+x^2f(x)
&=\sum_{n=2}^{\infty} a_{n-1}x^{n}+\sum_{n=3}^{\infty} a_{n-2}x^{n}\\
&=a_1x^2+\sum_{n=3}^{\infty} (a_{n-1}+a_{n-2})x^{n}\\
&=a_1x^2+\sum_{n=3}^{\infty} a_{n}x^{n}\\
&=a_1x^2+f(x)-a_1x-a_2x^2\\
\text{so}\\
f(x)(x^2+x-1)
&=(a_1-a_2)x^2-a_1x\\
\end{array}
$
Put in the initial
$a_1, a_2$
and $x = \frac14$.
Note that
this does not need
the explicit formula
for the $a_n$.
A: First, $a_{n}$ has the same definition as the Fibonacci numbers, so that part is easy enough.  
$a_{n} = \frac{\phi^n-\psi^n}{\sqrt{5}}$.  
https://en.wikipedia.org/wiki/Fibonacci_number
With a bit of re-arranging, you will see that its the difference of two geometric series.
$\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n+1}}=\frac{1}{4}\sum_{n=1}^{\infty} \frac{a_{n}}{4^{n}}=\frac{1}{4\sqrt{5}}\sum_{n=1}^{\infty} \frac{\phi^n-\psi^n}{4^{n}}=\frac{1}{4\sqrt{5}}\sum_{n=1}^{\infty} (\frac{\phi}{4})^n-\frac{1}{4\sqrt{5}}\sum_{n=1}^{\infty} (\frac{\psi}{4})^n$
https://en.wikipedia.org/wiki/Geometric_series
Plug and chug!
A: Hint $S=\sum\limits_{n=1}^{\infty}\dfrac{a_n}{4^{n+1}}=\frac 1{4\sqrt{5}}\sum\limits_{n=1}^{\infty}\left(\frac{\phi}4\right)^n-\left(\frac{\bar\phi}4\right)^n=\dfrac 4{11}$
A: 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!
