Prove that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ (suggestions for improvement on basic induction style/proof) Please see my proof to the following proposition below. From a standpoint of critiquing does this seem like a great execution or would someone feel that this proof was overly cumbersome? Specifically for the purposes of marks in a class how would this proof be viewed?
Prove that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ for $n\in\mathbb{N}$.
$Proof.$ We use mathematical induction.
Base Case. Let $n=1$. We see that $3^n=3^{(1)}=3$ and $\frac{3^{n+1}-3}{2}=\frac{3^{(1)+1}-3}{2}=\frac{6}{2}=3.$ Thus our equality holds for $n=1$.
Inductive Step. Let $n\ge1$. Assume $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$. Now observe that
$$\begin{align*}
3^1+3^2+3^3+3^4+\cdots+3^n+3^{n+1}&=\\
\frac{3^{n+1}-3}{2}+3^{n+1}&=\\
\frac{3^{n+1}-3+2(3^{n+1})}{2}&=\\
\frac{3^{n+1}+2(3^{n+1})-3}{2}&=\\
\frac{3^{n+1}(1+2)-3}{2}&=\\
\frac{3^{n+1}(3)-3}{2}&=\\
\frac{3^{n+2}-3}{2}&=\\
\frac{3^{(n+1)+1}-3}{2}&=\\
\end{align*}$$
It follows by induction that $3^1+3^2+3^3+3^4+\cdots+3^n=\frac{3^{n+1}-3}{2}$ for $\forall n\in\mathbb{N}$

Your proof is good!
Here is another nice way to prove it:
$$S=3^1+3^2+3^3+...+3^n\\3S=3(3^1+3^2+3^3+...+3^n)\\3S=3^2+3^3+3^4+...+3^{n+1}\\3S-S=3^{n+1}[-3^n+3^n]-...-[3^2+3^2]-3^1\\2S=3^{n+1}-3\\S=\frac{3^{n+1}-3}{2}$$
