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.

The formula must be a quadratic polynomial (because its first order difference is a linear polynomial) and has three independent coefficients. It suffices to identify for three different values of $n$:
$$\begin{align}1=\frac{1(3\cdot1-1)}2
\\1+4=\frac{2(3\cdot2-1)}2
\\1+4+7=\frac{3(3\cdot3-1)}2\end{align}$$
This completes the proof for any $n$ !
