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.
$\endgroup$ 44 Answers
$\begingroup$So, we've got $\dfrac{1}{-2^n \cdot -2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{1}{2^n \cdot 2^n} + \dfrac{1}{2^n \cdot 2^n} = \dfrac{2}{2^{2n}}$ since adding two of the same things up is the same as multiplying by $2$.
Alright, so you have $\frac{2}{2^{2n}}$. That's good work! Now use the property that $\frac{a}{b} = ab^{-1}$, since $\frac{1}{b} = b^{-1}$. Different notations, same thing.
Now, that means you have $\frac{2}{2^{2n}} = 2\cdot 2^{-2n}$. We're almost there: remember that $a^b \cdot a^c = a^{b+c}$, this can be seen by writing $a^b \cdot a^c = \underbrace{a \cdot a \cdots a}_{b \, \text{times}} \cdot \underbrace{a \cdot a \cdots a}_{c \, \text{times}} = \underbrace{a \cdot a \cdots a}_{(b+c) \, \text{times}} = a^{b+c}$.
Applying this to your case with $a = 2$, $b = 1$, $c = -2n$ we have $2 \cdot 2^{-2n} = 2^{1 - 2n} = 2^{-2n + 1}$.
$\endgroup$ 2 $\begingroup$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.
$\endgroup$ 4 $\begingroup$we have $$a^{-2}=\frac{1}{a^2}$$ thus we get $$(-2^n)^{-2}=\frac{1}{(-2^n)^{2}}=\frac{1}{2^{2n}}$$ further is $$(2^{-n})^2=\frac{1}{2^{2n}}$$
$\endgroup$ 1 $\begingroup$$(-2^n)^{-2} + (2^{-n})^2=$
$\frac 1{(-2^n)^2} + (\frac{1}{2^n})^2 = $
$\frac 1{(2^n)^2} + \frac 1 {(2^n)^2} = $
$\frac 1{2^{2n}} + \frac 1{2^{2n}} = $
$\frac 2{2^{2n}} = $
$\frac 1{2^{2n-1}} = $
$2^{-(2n-1)} = $
$2^{-2n+1}$
Or if you are more comfortable with the rules of exponents:
$(-a)^{b_{\text{even}}} = a^b$ so
$(-2^n)^{-2} + (2^{-n})^2 = (2^n)^{-2} + (2^{-n})^2 = 2^{-2n} + 2^{-2n} = 2*2^{-2n} = 2^{-2n + 1}$
$\endgroup$
