We define $E_n$ to be the number of ternary strings of length $n$ with an even sum and $O_n$ to be the number of ternary strings of length $n$ with an odd sum.
I think I found some expressions for $E_n$ and $O_n$ and they are as follows:
$$E_n = \frac{3^n + 1}{2}, \hspace{5mm} O_n = \frac{3^n - 1}{2}.$$
I found these just by testing some examples, but how would I prove this rigorously using induction.
Assuming that by "even/odd sum" you're referring to the sum of their digits...
Try looking at all the strings that contribute to $E_{n+1}$ and look at their first digit. Break it down by the three possible cases, and use your induction hypothesis to count the number of possibilities for the "tail" of each string, where by tail I mean "the length $n$ string you'd get by chopping off the first digit". And then do the same thing for $O_{n+1}$.
It might look a bit different from induction arguments you've done before, because you can't do $E_n$ and $O_n$ separately; your induction hypothesis will need to refer to the formulas for both of them.