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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 $\forall n\in\mathbb{N}$

ClownInTheMoon
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    Looks good. You probably don't need to show every step in the derivation of $\frac{3^{(n+1)+1}-3}{2}$. The first two or three and last line are probably enough. – Mosquite Sep 21 '16 at 20:42
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    In base-$3$ $$ 2\cdot 11111111 + 1 = 22222222+1 = 100000000 $$ It is just that. We are used to the base-$10$ equivalent. – Jack D'Aurizio Sep 21 '16 at 20:42

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Your proof is good!

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

76david76
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