Combinatorial Proof: How many length-n lists can we form using the elements in {1,2,3} [PROOF]

Question

I'm trying to prove that

$2\times(3^0) + 2\times(3^1) + 2\times(3^2) + \cdots+ 2\times(3^(n-1)) = 3^n - 1$ by answering the question "how many length-n lists can we form using the elements in $\{1,2,3\}$ in which the elements are not all 3 on both sides?" for both the LHS and RHS

So far, I assume that the RHS is $3^n - 1$ because it represents the number of lists of length $3$ minus the set of all $3'$s.

Any help would be appreciated! Thank you!

Answer 1

Hint: at every place in the list you can put one of $1,2,3$.