The question that is asked is to find the series described, and than calculate the sum of the first n terms.

Now I have done some research and I found a formula I thought might work, which is 
$$ s\_n = \frac{n}{2}(2a\_1 + (n-1)d)$$
where the series is of the form 
$a\_n = a\_1 + (n - 1)d$

Now I have got the following series:
$ 1 + 5 + 9 + 13$ $+$ $...$

This is what I though to be $s\_n$, so that $a\_1 = 1$, $a\_2 = 5$, etc.
I found the series description: $a\_n = a\_1 + 4(n-1)$ 

So I think $a\_1 = 1$ and $d = 4$.

When I substitute these values in the formula I get:

$$ s\_n = \frac{n}{2} (2 + 4n - 4) $$
$$ s\_n = \frac{2n + 4n^2 - 4n}{2} $$
$$ s\_n = 2n^2 - 2n $$

Yet, when I check the solution to the problem, it is expected to be 
$$ s\_n = n(2n - 1)$$
$$ s\_n = 2n^2 - n$$

I can't seem to figure out why I have almost the same result, yet a small difference.