<s>[INST] Is my proof, by strong induction, of for all $n\in\mathbb{N}$, $G_n=3^n-2^n$ correct? Let the sequence  $G_0, G _1, G_2, ...$ be defined recursively as follows: 
$G_0=0, G_1=1,$ and $G_n=5G_{n-1}-6G_{n-2}$ for every $n\in\mathbb{N}, n\ge2$. 
Prove that for all $n\in\mathbb{N}$, $G_n=3^n-2^n$.

Proof. By strong induction. Let the induction hypothesis, $P(n)$, be $G_n=3^n-2^n$ 
Base Case: For $(n=0)$, $P(0)$ is true because $3^0-2^0 =0$  
For $(n=1)$, $P(1)$ is true because $3^1-2^1=1$
Inductive Step: Assume that $P(n-1)$ and $P(n-2)$, where $n\ge2$, are true for purposes of induction.
So, we assume that $G_{n-1}=3^{n-1}-2^{n-1}$ and $G_{n-2}=3^{n-2}-2^{n-2}$, and we must show that $G_{ n }=3^{ n }-2^{ n }$.
Since we assumed $P(n-1)$ and $P(n-2)$, we can rewrite $G_n=5G_{n-1}-6G_{n-2}$ as $G_n=5(3^{n-1}-2^{n-1})-{ 6 }(3^{n-2}-2^{n-2})$
So, we get:
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-(\frac { 6 }{ 3 } \cdot 3^{ n-1 }-\frac { 6 }{ 2 } \cdot 2^{ n-1 })$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }-2\cdot 3^{ n-1 }+3\cdot 2^{ n-1 }$
$\Rightarrow G_n=5\cdot 3^{ n-1 }-2\cdot 3^{ n-1 }-5\cdot 2^{ n-1 }+3\cdot 2^{ n-1}$
$\Rightarrow G_n=3\cdot 3^{ n-1 }-2\cdot 2^{ n-1 }$
$\Rightarrow G_n=\frac { 1 }{ 3 } \cdot 3\cdot 3^n-\frac { 1 }{ 2 } \cdot 2\cdot 2^n$
$\Rightarrow G_n=3^n-2^n$

The only real issue I have at this point is that I don't know how to properly conclude this proof with a final statement. A hint/guidance in that regard would be much appreciated. 
In addition, please feel free to offer advice and/or constructive criticism about my proof. 
 [/INST] Yes, your proof is perfectly fine. Good job! You can write something like "The assertion follows.". But honestly it isn't necessary since it is in this case pretty simple for readers to see where the proof is complete (after the inductive step).
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