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.

Sum of the first and last terms = $1 + (3n-2) = 3n-1$
Sum of 2nd and (n-1)th terms =  $4 + (3n-5) = 3n-1$
Sum of 3rd and (n-2)th terms =  $7 + (3n-8) = 3n-1$
$...$
Sum of (n-1)th and 2nd terms =  $(3n-5) + 4 = 3n-1$
Sum of n-th (last) and 1st terms =  $(3n-2) + 1 = 3n-1$
Add both sides up . 
$(1+4+...+(3n-2)) + (1+4+...+(3n-2)) =  n(3n-1)$
which means:
$2(1+4+...+(3n-2)) =  n(3n-1)$
or
$1+4+...+(3n-2) =  n(3n-1)/2$
