<s>[INST] induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) + (n + 1)
\end{align*}
That seems way wrong lol ^^^
2nd attempt:
\begin{align*}
f(n) & = 3 + 9 + 15 + ... + (6n - 3)\\
f(n + 1) & = 6(n + 1) - 3\\
f(n + 1) & = 6(n - 3) + 6(n + 1) - 3\\
         & = ?
\end{align*} 
I don't know I feel like I'm headed in the wrong direction
I guess another attempt I have would be:
\begin{align*}
f(n) & = 3n^2\\
f(n+1) & = 3(n + 1)^2\\
       & = 3(n^2 + 2n + 1)\\
       & = 3n^2 + 6n + 3\\
       & = f(n) + (6n + 3)
\end{align*}
 [/INST] An other way
$$\sum_{k=1}^n(6k-3)=6\sum_{k=1}^n k-3n=6\cdot\frac{n(n+1)}{2}-3n=3n^2.$$
The only proof you need to do (by induction if you need to use induction) is that $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$
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