<s>[INST] 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.
 [/INST] The base case is trivial, now we follow to the inductive step by asuming the induction hypothesis and proving for $n + 1$.
So:
\begin{align*}
1 + 4 + 7 + ... + 3n-2 + 3(n+1)-2 & = \frac{n(3n-1)}{2} + 3(n+1)-2\\
                                  & = \frac{n(3n-1)}{2} + \frac{2(3(n+1)-2)}{2}\\
                                  & = \frac{3n^{2}-n+6n+6-4}{2}\\
                                  & = \frac{3n^{2}+5n+2}{2}\\
                                  & = \frac{(n+1)(3n+2)}{2}\\
                                  & = \frac{(n+1)(3(n+1)-1)}{2}
\end{align*}
And we are done. 
The important thing is to know when to apply the induction.
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