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 solution must be a quadratic polynomial (because its first order difference is a linear polynomial) so it is obtained with the Lagrangian formula on three points. Taking $(0,0),(1,1),(2,5)$,
$$0\frac{(n-1)(n-2)}{(0-1)(0-2)}+1\frac{(n-0)(n-2)}{(1-0)(1-2)}+5\frac{(n-0)(n-1)}{(2-0)(2-1)}=\frac{n(3n-1)}2.$$
