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 $n$
$\dfrac{3^k - 1}2 + 3^k=\dfrac32 3^k - \dfrac12=\dfrac12 3^{k+1} - \dfrac12=\dfrac12(3^{k+1} - 1)$