<s>[INST] Prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1)$ Using induction
prove that $1+4+7+...+(3n-2) = \frac{n}{2}(3n-1) \forall n \in \mathbb{N}$ 
Attempt:
Let $n =1$ so $3(1)-2 = 1$ and $\frac{1}{2}(3(1)-1)=1$
Assume true at $n=k$ so $3k-2 = \frac{k}{2}(3k-1)$
What do I do next? Here's where I'm stuck:
Let $n=k+1$ So $3(k+1) -2 = \frac{k+1}{2}(3(k+1)-1)$ 
 [/INST] Notice that you have an aritmetic progression with initial term $1$ and difference $3$. So, $a_1 = 1$, and $a_k = 1 + 3(k-1)$. Then, $3n - 2 = 1 + 3(k-1) \implies k = n$, so $3n - 2$ is the $n$th term of the progression. This way: $$S_n = n\frac{a_1 + a_n}{2} = \frac{n}{2}(3n-1).$$
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