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*}
 A: 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 $$\sum_{k=1}^nk=\frac{n(n+1)}{2}.$$
A: Using summation notation gives
$$3 + 9 + 15 + \cdots + 6n - 3 = \sum_{k = 1}^n (6k - 3)$$
Let $P(n)$ be the statement 
$$\sum_{k = 1}^{n} (6k - 3) = 3n^2$$
To prove this statement by induction, we must show that $P(1)$ holds and that whenever $P(m)$ holds for some positive integer $m$, then $P(m + 1)$ holds since we then have the chain of implications 
$$P(1) \implies P(2) \implies P(3) \implies \cdots$$
Proof.  Let $n = 1$.  Then 
$$\sum_{k = 1}^{1} (6k - 3) = 6 \cdot 1 - 3 = 6 - 3 = 3 = 3 \cdot 1 = 3 \cdot 1^2$$
so $P(1)$ holds.
Since $P(1)$ holds, we may assume there exists a positive integer $m$ such that $P(m)$ holds.  Then 
$$\color{green}{\sum_{k = 1}^{m} (6k - 3) = 3m^2}$$
Let $n = m + 1$.  Then
\begin{align*}
\sum_{k = 1}^{m + 1} (6k - 3) & = \color{green}{\sum_{k = 1}^{m} (6k - 3)} + [6(m + 1) - 3] && \text{by definition}\\
                              & = \color{green}{3m^2} + 6(m + 1) - 3 && \text{by the induction hypothesis}\\
                              & = 3m^2 + 6m + 6 - 3\\
                              & = 3m^2 + 6m + 3\\
                              & = 3(m^2 + 2m + 1)\\
                              & = 3(m + 1)^2
\end{align*}
Hence, $P(m) \implies P(m + 1)$.  
Since $P(1)$ holds and $P(m) \implies P(m + 1)$ for each positive integer $m$, $P(n)$ holds for all positive integers.$\blacksquare$
A: Without induction you can do it easily taking three common so $3(1+3+..(2n-1))$ now we know that sum of odd terms is a perfect square so we can write as $3n^2$ or sum of AP=$\frac{n}{2}(2+(n-1)2)=n^2$
A: If you insist on induction :
Base case $n=1\ :\ 6\times 1-3=3=3\times 1^2$
If you assume $$3+9+15+...+(6n-3)=3n^2$$
You can conclude
$$3+9+15+...+(6n-3)+(6n+3)=3n^2+6n+3=3(n+1)^2$$ 
completing the proof.
