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.

Here is @Shooter's answer shown a different way.  Let's take the example of n = 8:
1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 = 92

Let's rearrange this and group:
(1 + 22) + (4 + 19) + (7 + 16) + (10 + 13) = 92

Now let's add the groups and look for a pattern:
23 + 23 + 23 + 23 = 92

That's it.  The n = 8 example is just what happens when you add 23 four times.  What is 23?  It's 3n - 1.  What is four?  It's n / 2.
That makes the formula (3n - 1)(n / 2).  This is how I would derive this when n is even.  It's a little more work when n is odd.

Here's another simple solution that only uses this formula:
1 + 2 + 3 + ... + n = n * (n + 1) / 2

We'll call this f(n) for now.  I'll decompose the original series:
(a)   1 + 2 + 3 + 4 ...
(b) + 0 + 1 + 2 + 3 ...
(c) + 0 + 1 + 2 + 3 ...
-------------------
(d) = 1 + 4 + 7 + 10 ...

It should be clear from the above how to make a closed form equation:
(a) = f(n)
(b) = f(n - 1)
(c) = f(n - 1)

Therefore
(d) = f(n) + 2 * f(n - 1)

The formula (d) can be rewritten to what you posted in your question
