Show that $3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other. 
$3^{-n}$ have the interesting property that one half of their repeating binary string is the inverse of the other.  Prove it!

$3^{-1}=\overline{0\color{red}{1}}_2$
$3^{-2}=\overline{000\color{red}{111}}_2$
$3^{-3}=\overline{000010010\color{red}{111101101}}_2$
$\ldots$
These are the binary representations - add a minus sign to the left hand side and the right hand side is the 2-adic representation.
 A: Lemma: For any sequence of $k$ digits $A$, where $B$ is the opposite sequence of digits: $$0.\overline{AB}_2 = \frac{A + 1}{2^k + 1}.$$
Proof: Let $x = 0.\overline{AB}$. Then
$$
2^{k} x = A.\overline{BA}.
$$
Therefore,
$$
x + 2^k x = A.\overline{11\ldots 1} = A + 1 \quad \implies \quad x = \frac{A + 1}{2^k + 1}.
$$

So we have to show that $3^{-n} = \frac{1}{3^n}$ has this form, $\frac{A+1}{2^k + 1}$, where $A$ is between $0$ and $2^{k} - 1$. Because $3^{-n}$ is definitely between $0$ and $1$, this is equivalent to just finding $k$ such that $3^n \mid 2^{k} + 1$.
From the pattern in the examples in your question, we guess that we should pick
$$
k = 3^{n-1}.
$$
So we show by induction that
$$
3^n \mid 2^{3^{n-1}} + 1.
$$
Base case: $3^1 \mid 2 + 1 = 3$.
Inductive step:
Let's assume that $3^n \mid 2^{3^{n-1}} + 1$, for some $n$. Specifically, let
$2^{3^{n-1}} = a \cdot 3^n - 1$.
 Then
\begin{align*}
2^{3^n} + 1
  &= \left(2^{3^{n-1}}\right)^3 + 1 \\
  &= \left(a \cdot 3^n - 1\right)^3 + 1 \\
  &= a^3 3^{3n} - 3 a^2 3^{2n} + 3 a 3^n - 1 + 1 \\
  &\equiv 0 - 0 + 0 - 1 + 1 \pmod{3^{n+1}} \\
  &= 0.
\end{align*}
So $2^{3^n} + 1$ is divisible by $3^{n+1}$, and the induction is complete.
A: In general, the period of $2$ in $\Bbb Z/(3^{N+1})^\times$ is $2\cdot3^N$. And we expect that the $2$-adic expansion of $-3^{N+1}$ should be purely periodic, period $2\cdot3^N$.
Indeed, since $3^{N+1}|(2^{2\cdot3^N}-1)$, say with quotient $Q_N$, we get the results
\begin{align}
Q_N&=\frac{2^{2\cdot3^N}-1}{3^{N+1}}\\
-\frac1{3^{N+1}}&=\frac{Q_N}{1-2^{2\cdot3^n}}\,,
\end{align}
in which the second line says that the number of binary digits in the repeating block of the $2$-adic expansion of $-3^{-N-1}$ is $2\cdot3^N$, and what’s in the block is the number $q_N$.
What we know is that $2^{2\cdot3^N}-1\equiv0\pmod{3^{N+1}}$, so we can factor
$$ \left(2^{3^N}-1\right)\left(2^{3^N}+1\right)\equiv0\pmod{3^{N+1}}\,, $$
but please note that since $3^{N+1}$ is odd, we see that the left-hand factor above is $\equiv1\pmod3$, in particular relatively prime to $3$, and thus to $3^{N+1}$ as well. Thus $3^{N+1}$ divides the right-hand factor, i.e. $3^{N+1}\mid(2^{3^N}+1)$, and once again to make typing easier for
myself, I’ll call the quotient $\Omega$. Thus we have:
\begin{align}
\Omega&=\frac{2^{3^N}+1}{3^{N+1}}\\
0&<\Omega<2^{3^N}\\
Q_N&=\Omega\left(2^{3^N}-1\right)\\
&=2^{3^N}(\Omega-1)+\left(2^{3^N}-\Omega\right)\\
\text{where we note }0&<2^{3^N}-\Omega<2^{3^N}\,.
\end{align}
And that gives us our expression for $Q_N=2^{3^N}a+b$ with both $a$ and $b$ in the interval $\langle0,2^{3^N}\rangle$, namely $a=\Omega-1$ and $b=2^{3^N}-\Omega$. And surenough, $a+b=2^{3^N}-1$, as we desired.
