How would I go about proving the following:

For any odd positive integer $s$, there exists a sequence of nonnegative integers $( a_0, a_1, \cdots, a_{n-1})$ and a nonnegative integer $m$ such that,

$$ 3^ns + \sum_{k=0}^{n-1} 3^{n-k-1}2^{a_k}=2^m.$$


I am really stuck. I was thinking of using the ring $\mathbb{Z}_{2^m}$ in a proof by contradiction, but I cannot even get started reducing the LHS to something simpler.


**Note 1:** I have reason to believe there exists such a sequence where 

 1. $a_0=0$
 3. $\{a_n\}$ is strictly monotonically increasing

**Note 2:** I think an example might help. 



$$ n=3 \quad \{a_n\} = [0, 2g-1, 2g+3] \quad s = \frac{5\times 4^g -2}{6} \quad m= 2g+5 \quad g >0$$


**Note 3:** I originally asked this on the [Mathematics Stack Exchange](https://math.stackexchange.com/questions/3414613/proof-of-3ns-sum-k-0n-1-3n-k-12a-k-2m), but it seems to be a question more suited for this exchange.