Answer
$f(n) == \frac{1000}{\sqrt{\sqrt{2}}}\cdot\left( \frac{1}{\sqrt{2}}\right)^n$
Work Step by Step
\begin{equation}
\begin{aligned}
f(n)&=1000\cdot 2^{-\frac{1}{4}-\frac{n}{2}}\\
&=1000\cdot 2^{-\frac{1}{4}}\cdot 2^{-\frac{n}{2}}\\
&= \frac{1000}{2^{\frac{1}{4}}}\cdot\left( \frac{1}{2^{\frac{1}{2}}}\right)^n\\
&= 500\sqrt[4]8\cdot\left( \frac{1}{\sqrt{2}}\right)^n
\end{aligned}
\end{equation}
Hence, the relationship between $a$ and $b$ of an A-series paper can be written as.
\begin{equation}
\begin{aligned}
b&=\frac{1}{\sqrt{2}}\\
a&=\frac{1000}{\sqrt{\sqrt{2}}}\\
&= 1000\sqrt{b}
\end{aligned}
\end{equation}
