What is the asymptotic behaviour of a sum of powers of three?
First, your assumption, I'm more concerned with just finding big oh, since that can be used to find big omega is a bit dangerous. You could indeed easily find a $O()$ upper bound for your sum, $S(n)=3^1+3^2+3^3+\cdots+3^n$ by observing that each term is less than $3^n$ so you could have 
$$
3^1+3^2+\cdots+3^n\le \underbrace{3^n+3^n+\cdots+3^n}_{n}=n3^n=O(n3^n)
$$
but that's a huge overestimate (try it for, say, $n=5$) and so will almost certainly not serve as a suitable big-Omega bound*. Instead, you could simply use the formula for a geometric series and get an exact value, but suppose you didn't know that formula (or were prohibited from simply invoking it).

On the way to finding a good estimate, you could always try this helpful idiom: We've written $S(n)=3^1+3^2+3^3+\cdots+3^n$ and so we also have $S(n+1)=3^1+3^2+3^3+\cdots+3^{n+1}$ so
$$\begin{align}
S(n+1)-3S(n)&amp;=(3^1+3^2+\cdots+3^{n+1})-3(3^1+3^2+\cdots+3^n)\\
 &amp;=(3^1+3^2+\cdots+3^{n+1})-(3^2+3^3+\cdots++3^{n+1})\\
 &amp;= 3
\end{align}$$
But we also know that $S(n+1)=S(n)+3^{n+1}$ so we have
$$
3=S(n+1)-3S(n)=S(n)+3^{n+1}-3S(n)=-2S(n)+3^{n+1}
$$
and so with a little algebra we find
$$
S(n) = \frac{3^{n+1}-3}{2}=\frac{3}{2}(3^n-1)
$$
Since we have an exact value for the sum, we can easily see that $S(n)=\Theta(3^n)$. However, if you need to establish explicit constants to show $S(n)\ge c\,3^n$ (for big-omega) and $S(n)\le d\,3^n$ (for big-o) it's not too difficult. (Hint, can you see that you can use any $d\ge 3/2$?)



$^*$As an aside, you might have been misled in your choice of replacements at the very start by the proof that $1^k+2^k+3^k+\cdots+n^k\le n^k+n^k+\cdots+n^k=n^{k+1}$ which does indeed lead to a big-Theta estimate, though for a different sum.