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: 
**Problem.** $\quad$Prove by induction the following formula: $\quad 1+3+3^2+\dots+3^{n−1} = \dfrac{1}{2}\big(3^n−1\big)\,$.

**Solution.** $\quad$Proof by induction is given in three following steps:

1. *Base:* assume $\,n=2,\,$ then $\;1+3 = 4 = \dfrac{1}{2}\big(3^2−1\big), \,$ so the formula is correct.
2. *Hypothesis:* assume formula holds for some positive integer $\,n.\,$
3. *Induction:* show that formula holds for $\,n+1,\;$ i.e. $\; 1+3+\dots+3^{n−1} + \color{#df0000}{3^n}= \dfrac{1}{2}\big(3^{n+1}−1\big)\,$

	* by induction hypothesis we have $\quad 1+3+3^2+\dots+3^{n−1} = \dfrac{1}{2}\big(3^n−1\big)\,$
	* let us add $\,3^n\,$ to the formula above in order to complete the induction step:
	\begin{align}
	\Big(1+3+3^2+\dots+3^{n−1}\Big) + \color{#df0000}{3^n} & = \dfrac{1}{2}\big(3^{n}−1\big) + \color{#df0000}{3^n} \\ & =
	\dfrac{1}{2}\big(3^{n}−1+2\cdot \color{#cc0000}{3^n}\big) \\ & =
	\dfrac{1}{2}\big(3\cdot \color{#df0000}{3^n} -1\big) \\ & =
	\dfrac{1}{2}\big(\color{#df0000}{3^{n+1}} -1\big).
	\end{align}Thus we established that 
$$
\bbox[4pt, border:solid 2pt #FC0000]{\;1+3+3^2+\dots+3^{n−1} + 3^{n} = \dfrac{1}{2}\big(3^{n+1}−1\big)}
$$

*Q.E.D.*
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