<s>[INST] 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?
 [/INST] 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: $\frac{3^{n+1}-1}{2}+3^{n+1}=\frac{3^{n+1}-1+2(3^{n+1})}{2}=\frac{3(3^{n+1})-1}{2}=\frac{3^1(3^{n+1})-1}{2}=\frac{3^{(n+1)+1}-1}{2}.$
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