<s>[INST] How do I prove that $\frac{(3^k-1)+6k}{2} $ is the same as $\frac{3^{k+1}-1}{2}$? Use mathematical induction to prove that the following statement is true for every positive integer $n$  
$1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}$
Here are my steps:


*

*Show that $S_1$ is true   


$S_1: 1=\frac{3^{(1)}-1}{2}$   
$S_1: 1=\frac{3 -1}{2}$  
$S_1: 1=\frac{2}{2}$  
$S_1: 1=1$


*Show that if $S_k$ is assumed to be true, then $S_{k+1}$ is also true, for every positive integer k.  


$S_k:1+3+3^2+...+3^{k-1}=\frac{3^k-1}{2}$  
$S_{k+1}:1+3+3^2+...+3^{(k+1)-1}=\frac{3^{k+1}-1}{2}$
So I begin by adding $3^{(k+1)-1}$ to both sides of $S_k$, and then simplifying from that point on until the final result is the statement $S_{k+1}$.
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{3^k-1}{2} + 3^{(k+1)-1}$  
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{3^k-1}{2} + 3^{k}$  
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{3^k-1}{2} + \frac{6^{k}}{2}$  
$S_k:1+3+3^2+...+3^{k-1}+3^{(k+1)-1}=\frac{(3^k-1)+6^k}{2}$   
But this is where I get stuck...
How do I prove that $\frac{(3^k-1)+6k}{2} $  is the same as $\frac{3^{k+1}-1}{2}$?
 [/INST] Proposition: $$1+3+3^2+...+3^{n-1}=\frac{3^n-1}{2}$$
Assume true for $n$
Now we must prove true for $n+1$:
$$1+3+3^2+...+3^{n-1}+3^n = \frac{3^n-1}{2} + 3^n = \frac{3^n-1+2\cdot 3^n}{2}$$
$$= \frac{3\cdot 3^n-1}{2} = \frac{3^{n+1}-1}{2}$$ as required, hence true $\forall n\in\mathbb{Z^+}$
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