<s>[INST] 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 $a_{1}=a_{2}=1$?
 [/INST] $a_{n+2}=a_{n+1}+a_{n}$
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 $a_n$.
</s>