How would I show this problem through mathematical induction? I am trying to learn a lot of this on my own but I have never tried proving something through mathematical induction. Here is the problem below.

$$1+3+3^2 + \cdots + 3^n = \frac{3^{(n+1)}-1}{2}$$
for all $n\in\mathbb{N}_0$, using mathematical induction.
  Note that here $\mathbb{N}_0$ means all integers $n \geq 0$.

I need to start with the basis step which would be:
If $n = 0$ then $3^0 = 1$, and $1 = 3^{0+1} = 3 - 1 = 2$, and then $2/2 = 1$. So I have proved the basis step but how do I do the inductive step for this?
 A: You say that you only need help with the inductive step. For simplicity, set $n = k$, where $k \geq 0$. We have that
$$1+3+3^2 + ... + 3^k = \frac{3^{k+1}-1}{2}$$
Now, let's look at the same scenario, except that we add the next part of the sequence, thus we are calculating for $k+1$:
$$1+3+3^2 + ... + 3^k+3^{k+1} = \frac{3^{k+2}-1}{2}$$
Note that we already have a formula for the first parts of the sequence (up to and including $3^k$):
$$\frac{3^{k+1}-1}{2} + 3^{k+1} = \frac{3^{k+2}-1}{2}$$
If you are able to show that this last statement is true, you are done.
A: This is call perturbation method, from $\mathit{Concrete \ Mathematics}$ by Graham, Knuth and Patashnik. Denote $S_n=\sum_{k=0}^{n}3^k$. You get:
$$
S_n = \sum_{k=0}^{n}3^k\\
S_n + 3^{n+1}=\sum_{k=0}^{n}3^k +3^{n+1}\\
S_{n}+ 3^{n+1} = 1+3\sum_{k=0}^{n}3^k\\
\text{a bit of algebra here}\\
S_n=\frac{3^{n+1}-1}{2}
$$
A: For the inductive step, assume your claim holds for $n$ and then try to prove it for $n+1.$ That is,
Assume: $1+3+3^2+...+3^n=\frac{3^{n+1}-1}{2}$.
Then see if you can prove that $1+3+3^2+...+3^n+3^{n+1}=\frac{3^{(n+1)+1}-1}{2}$.
How to do this? Start by transforming the left hand side using what you already know:
$1+3+3^2+...+3^n+3^{n+1}=\frac{3^{n+1}-1}{2}+3^{n+1}.$
Now combine these fractions and see if you can massage them into $\frac{3^{(n+1)+1}-1}{2}$. 
That is: $\frac{3^{n+1}-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2(3^{n+1})}{2}=\frac{3(3^{n+1})-1}{2}=\frac{3^1(3^{n+1})-1}{2}=\frac{3^{(n+1)+1}-1}{2}.$
