If $n$ is a positive integer, then $(-2^n)^{-2} + (2^{-n})^2 = 2^{-2n+1}$ I'm not sure why $$(-2^n)^{-2} + (2^{-n})^2=2^{-2n+1}$$
I have been going over this equation for a while now, noticing, and have successfully got quite far in the equation, finding that 
$$ (-2^n)^{-2} + (2^{-n})^2 \implies \frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} $$
which I think then becomes $ \dfrac{2}{2^{2n}}$
But then I get stuck.

The right hand side of the implies sign above can be studied, viz;
\begin{align}
\frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} &= \frac{2^{2n}+(-1)^{2n}2^{2n}}{(-1)^{2n}2^{4n}}\\
&= \frac{2.2^{2n}}{2^{4n}} \qquad (-1)^{2n}=1 \quad \forall n\\
&= \frac{2}{2^{2n}}
\end{align}
Thus,
\begin{align}
\frac{1}{(-2^n)^2} + \frac{1}{(2^n)^2} &= \frac{2}{2^{2n}} \\
\implies (-2^n)^{-2} + (2^n)^{-2}&=2.2^{-2n} \\
&= 2^{-2n+1}
\end{align}
Hope that helps.
