Prove that $1 + 4 + 7 + · · · + 3n − 2 = \frac {n(3n − 1)}{2}$ Prove that
$$1 + 4 + 7 + · · · + 3n − 2 =
\frac{n(3n − 1)}
2$$
for all positive integers $n$.

Proof: $$1+4+7+\ldots +3(k+1)-2= \frac{(k + 1)[3(k+1)+1]}2$$
  $$\frac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$$

Along my proof I am stuck at the above section where it would be shown that:

$\dfrac{(k + 1)[3(k+1)+1]}2 + 3(k+1)-2$   is equivalent to $\dfrac{(k + 1)[3(k+1)+1]}2$
Any assistance would be appreciated.

Non-inductive derivation:
\begin{align}
\sum_{k=1}^n(3k-2) &= \sum_{k=1}^n3k -\sum_{k=1}^n2\\
&= 3\left(\sum_{k=1}^n k\right) -2n\\
&= \frac{3(n)(n+1)}{2} - \frac{4n}{2}\\
&=\frac{3n^2-n}{2}\\
&= \frac{n(3n-1)}{2}\\
\end{align}
This, of course, relies on one knowing the sum of the first $n$ natural numbers, but that's a well-known identity.
