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.

Isn't this a super-standard arithmetic progression? https://en.wikipedia.org/wiki/Arithmetic_progression
In your case:
first = element(1) = 1
element(N) = 3N-2
Sum of N first elements of arithmetic progression is N(first+element(N))/2 which is...
N(1+ 3N-2)/2
What's to prove here? Or are you trying to prove the general equation for a sum of first N elements of arithmetic progression? 
