Answer
$g(n)=1000(0.7071)^n$
Work Step by Step
We have $g(n)=\sqrt{f(n) \cdot f(n-1)}$ where $$
\begin{aligned}
f(n) & =1000 \cdot 2^{-\frac{1}{4}-\frac{n}{2}} \\
& =1000 \cdot 2^{-\frac{1}{4}} \cdot 2^{-\frac{n}{2}}
\end{aligned}
$$ and $$
\begin{aligned}
f(n-1) & =1000 \cdot 2^{-\frac{1}{4}-\frac{n-1}{2}} \\
& =1000 \cdot 2^{-\frac{1}{4}} \cdot 2^{-\frac{n-1}{2}} \\
& =1000 \cdot 2^{-\frac{1}{4}} \cdot 2^{\frac{1-n}{2}} \\
& =1000 \cdot 2^{-\frac{1}{4}} \cdot 2^{\frac{1}{2}-\frac{n}{2}} \\
& =1000 \cdot 2^{-\frac{1}{4}} \cdot 2^{\frac{1}{2}} \cdot 2^{-\frac{n}{2}} \\
& =1000 \cdot 2^{-\frac{1}{4}+\frac{1}{2}} \cdot 2^{-\frac{n}{2}} \\
& =1000 \cdot 2^{\frac{1}{4}} \cdot 2^{-\frac{n}{2}}
\end{aligned}
$$ So that $$
\begin{aligned}
f(n) \cdot f(n-1) & =\left(1000 \cdot 2^{-\frac{1}{4}} 2^{-\frac{n}{2}}\right)\left(1000 \cdot 2^{\frac{1}{4}} \cdot 2^{-\frac{n}{2}}\right) \\
& =1000 \cdot 1000 \cdot 2^{-\frac{1}{4}} \cdot 2^{\frac{1}{4}} \cdot 2^{-\frac{n}{2}} \cdot 2^{-\frac{n}{2}} \\
& =1,000,000 \cdot 2^{-n}
\end{aligned}
$$ Hence $$
\begin{aligned}
g(n) & =\sqrt{f(n) \cdot f(n-1)} \\
& =\sqrt{1,000,000 \cdot 2^{-n}} \\
& =\sqrt{1,000,000}\cdot \sqrt{2^{-n}} \\
& =1000\left(2^{-n}\right)^{0.5} \\
& =1000 \cdot 2^{-0.5 n} \\
& =1000\left(2^{-0.5}\right)^n \\
& =1000(0.7071)^n .
\end{aligned}
$$
