Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$ This is the assignment I have:

Find a closed form for the equations 
$1^3 = 1$
$2^3 = 3+5$
$3^3 = 7+9+11$
$4^3 = 13+15+17+19$
$5^3 = 21+23+25+27+29$
$...$
Hints. The equations are of the form $n^3 = a1 +a2 +···+an$, where
  $a_{i+1} = a_i +2$ and $a_0 =n(n−1)+1$.

My reasoning:
We have to find a formula that give us $n^3$ summing operands. (why is this useful?)
We know that the first operand (or term) of the sum is $a_0 =n(n−1)+1$.
In fact, if you put $n = 3$, then $a_0 = 3(3 − 1) + 1 = 3*2 + 1 = 7$, which is exactly the first number of sum.
Then I notice that each $n$ sum has $n$ operands, and each operand differs from one another of 2.
Thus I came out with this formula:
$$
\sum\limits_{i=0}^{n-1} a_0 + 2 \cdot i
$$
where $a_0 =n(n−1)+1$
For example, if $n = 3$, then we have 
$(n(n−1)+1 + 2 \cdot 0) + (n(n−1)+1 + 2 \cdot 1) + (n(n−1)+1 + 2 \cdot 2) \equiv$
$\equiv (7 + 0) + (7 + 2) + (7 + 4) \equiv$
$\equiv 7 + 9 + 11$
Which is what is written as third example.
I don't know if this is correct form or even if this is a closed form, that's why I am asking...

$$n^3=\sum_{k=0}^{n-1}(n^2-(n-1)+2k)$$
Since
$\sum_{k=0}^{n-1}(n^2-(n-1)+2k)=n^3-n(n-1)+2(\frac{n(n-1)}{2})=n^3$
So it is the summation of $n$ consequitive odd number.
