$3$-adic expansion of $- \frac{9}{16}$ I get the $3$-adic expansion to be $1+1 \cdot 3+2 \cdot 3^2 +2 \cdot 3^3 + 0 \cdot 3^4+\cdots$. I'm trying to work out a pattern of the coefficients and think it is $1, 1, 2, 2, 0, 0, 1, 1, 2, 2, 0, 0,...$. To show this I need to show that $1+1 \cdot 3+2 \cdot 3^2 +2 \cdot 3^3 + 0 \cdot 3^4 + \cdots$ sums to $- \frac{9}{16}$, but I don't know how to manipulate this series in order to compute it.
Any help would be appreciated!

I will explain the case of $3$ - adic of $\frac{21}{50}$ and leave it to you to solve for $\frac{-9}{16}$
For $\frac{21}{50}$ you have expression $\sum_{i\in \mathbb{Z}} a_i p^i$ : $0\leq a_k<p$
$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots$$
So, how do you get $a_0$ ???
$a_0$ is chosen such that
$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots$$
i.e., $$\frac{21}{50}-a_0=a_1p+a_2p^2+a_3p^3+\cdots$$
see that $p$ divides $a_1p+a_2p^2+a_3p^3+\cdots$ so $p$ should divide $\dfrac{21}{50}-a_0$
i.e., $\dfrac{21}{50}-a_0\equiv 0 ~\text{mod} ~3$ i.e., $3$ divides $21-50a_0$
You have only three possibilities for $a_0$ namely $0,1,2$... Check it..
Now,$a_1$... $a_1$ is chosen such that
$$\frac{21}{50}=a_0+a_1p+a_2p^2+a_3p^3+\cdots\Rightarrow\frac{21}{50}-a_0-a_1p=a_2p^2+a_3p^3+\cdots$$
see that $p^2$ divides $a_2p^2+a_3p^3+\cdots$ so $p^2$ should divide $\dfrac{21}{50}-a_0-a_1p$
You know what is  $a_0$. substitute it in above expression..
$\dfrac{21}{50}-a_0-3a_1\equiv 0 ~\text{mod} ~3^2$ 
Just repeat this accordingly and you will find what other $a_k$ are...
