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...
 A: $$\begin{align}
1^3&=1\\
2^3&=3+5\\
3^3&=7+9+11\\
4^3&=13+15+17+19\\
5^3&=21+23+25+27+29\\
\vdots &= \vdots\\
n^3&=[(n^2-n+1)]+[(n^2-n+1)+2]+[(n^2-n+1)+4]+\cdots+[(n^2-n+1)+2(n-1)]\\
&=\sum_{r=1}^n(n^2-n+1)+2(r-1)\qquad \blacksquare
\end{align}$$
That appears to be the formula required for the "series". 
The formula expresses the cube of an integer ($n$) as the sum of $n$ integers which are  in arithmetic progression with common difference of $2$ (and is not so much about the summation of the first $n$ cubes).
To show that this is correct:
$$\begin{align}
\text{RHS}&=\sum_{r=1}^n[\color{blue}{(n^2-n+1)}+2(r\color{blue}{-1})]\\
&=n(n^2-n-1)+2\sum_{r=1}^n r\\
&=n(n^2-n-1)+2\cdot \frac {n(n+1)}2\\
&=n^3=\text{LHS}\qquad \blacksquare
\end{align}$$
A: Let us derive  the $r$ term of $1,3,7,13,21$ 
Let $S_r=1+3+7+13+21+\cdots+T_r$
$S_r-S_r=1+(3-1)+(7-3)+(13-7)+(21-13)+\cdots+T_r-T_{r-1}-T_r$
$\implies T_r=1+(2+4+6+8+$ up to $r-1$th term $)$
$=1+\dfrac{(r-1)}2\{2\cdot2+(r-2)2\}=1+r^2-r$
Now the $m$th row will be, $$\sum_{r=0}^{m-1}\{(m^2-m+1)+2r\}=(m^2-m+1)\sum_{r=0}^{m-1}1+2\sum_{r=0}^{m-1}r$$
$$=m(m^2-m+1)+2\cdot\frac{m(m-1)}2=m^3$$
A: $$n^3=\sum_{k=0}^{n-1}(n^2-(n-1)+2k)$$
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.
