Induction Proof $2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{ 2}$ I am looking for an induction proof...
$$2 + 5 + 8 + 11 + \cdots + (9n - 1) = \frac{3n(9n + 1)}{2}$$ 
when $n \geq 1$.
I know there are $3$ steps to this.
1) Check
2) Do $n = k$
3) Do $n = k + 1$
Problem is, I can't seem to get past the first step, when I check:
$n = 1: (9(1) - 1) = 8$ but isn't it suppose to be come to $2$?

For $n=1$, Ok. It's easy to check
Now assume it works for $n$, the we have to show it works for $n+1$.
(note that $9(n+1)-1=9n+8$)
we have to show the sentence below is true:
$$2+5+8+...+9n-1+9n+2+9n+5+9n+8=\frac{3(n+1)(9(n+1)+1)}{2}=\frac{27n^2+27n+30}{2}$$
Knowing it works for $n$, then we have:
$$2+5+8+...+9n-1+9n+2+9n+5+9n+8$$
$$=\frac{3n(9n+1)}{2}+9n+2+9n+5+9n+8$$
$$=\frac{27n^2+3n}{2}+27n+15$$
$$=\frac{27n^2+57n+30}{2}$$ $$=\frac{3(n+1)(9(n+1)+1)}{2}.$$
