Question: > 
> $1 + 3 + 3^2 + ... + 3^{n-1} = \dfrac{3^n - 1}2$
> 
> 
> 

I am stuck at

$\dfrac{3^k - 1}2 + 3^k$

and I'm not sure if I am right or not.

Answer: 
For $n=1$ is true because $1=(3^1-1)/2=2/2=1$.
Suppose that is true for $n\leq k$.
$1+...+3^{k-1}+3^k=\_{HI} \frac{3^k-1}{2}+3^k=\frac{3}{2}3^k-\frac{1}{2}=\frac{3^{k+1}}{2}-\frac{1}{2}=\frac{3^{k+1}-1}{2}$. And then is true for all $n$
Assume
$$S=1+3+3^2+3^3+...+3^{n-1}$$
in this case
$$ 3\cdot S=3\cdot1+3\cdot3+3\cdot3^2+...+3\cdot3^{n-2}+3\cdot3^{n-1}$$
$$ 3\cdot S=3+3^2+3^3+...+3^{n-1}+3^{n}$$
Now subtracting $S$ from $3\cdot S$ we have:
$$3\cdot S-S=-1+3-3+3^2-3^2+3^3-3^3+...+3^{n-1}-3^{n-1}+3^n$$
it is easy to see that the result will be
$$2\cdot S=3^n-1$$
so
$$S=\frac{3^n-1}{2}$$
Replacing $S$ with the original sum we have:
$$1+3+3^2+3^3+...+3^{n-1}=\frac{3^n-1}{2}$$
Q.E.D