Cubes as the sum of odd integers It is well known that
$1^3=1$
$2^3=3+5$
$3^3=7+9+11$
$4^3=13+15+17+19$
$5^3=21+23+25+27+29$
and so on. This is typically proven using induction. I have come up with a proof and I'm wondering what you guys think or if you have seen this solution before :)
We will consider the array
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
&&&&\ddots
\end{matrix}
\end{align*}
and in the fashion of matrices, we let $A_{ij}$ denote the entry in row $i$ and column $j$. To be clear, $A_{11}=1, A_{21}=3, A_{22}=5$, etc. Then it suffices to show that $\sum_{j=1}^i A_{ij}=i^3$. Let us consider our array up to row $i$.
\begin{align*}
\begin{matrix}
1\\
3 & 5\\
7 & 9 & 11\\
13& 15 & 17 & 19\\
\vdots \\
A_{(i-1)1}&...&A_{(i-1)(i-1)}\\
A_{i1}&...&A_{ij} &...&A_{ii}
\end{matrix}
\end{align*}
It is clear to see that for $i \geq 2$ we have $A_{ii}=A_{(i-1)(i-1)}+2i$ as row $i$ consists of the $i$ odds following $A_{(i-1)(i-1)}$. We can solve for $A_{(i-1)(i-1)}$ by iteration.
\begin{align*}
A_{(i-1)(i-1)}&=A_{(i-2)(i-2)}+2(i-1)\\
&=A_{(i-3)(i-3)}+2(i-1)+2(i-3)\\
&=A_{(i-4)(i-4)}+2(i-1)+2(i-3)+2(i-4)\\
&...\\
&=1+2(i-1)+2(i-3)+2(i-4)+...+2(3)+2(2)\\
&=(i-1)i-1.
\end{align*}
Remarking that $A_{ij}=A_{(i-1)(i-1)}+2j$, we conclude that $A_{ij}=(i-1)i-1+2j$. Making use of this formula, it follows that $\sum_{j=1}^i A_{ij}=i^3$ as desired.
Let me know if there is any clarification necessary!
 A: See my answer to this question.
It is a little appreciated fact that every power $k\ge 2$ of any positive integer $n$ can be expressed as the sum of exactly $n$ consecutive odd numbers, viz:  $$n^k=\sum_{i=\frac{n^{k-1}-n}{2}+1}^{\frac{n^{k-1}+n}{2}}(2i-1)$$
So $n$ consecutive odd numbers can be found that sum to $n^3$ for any $n$.
$$n^3=\sum_{i=\frac{n^{2}-n}{2}+1}^{\frac{n^{2}+n}{2}}(2i-1)$$
If you compute the starting and ending values for the summation for any particular $n$, you get exactly the numbers in your exposition.
The general formula specifies $n$ consecutive odd numbers with an average value of $n^{k-1}$ summing to $n\cdot n^{k-1}=n^k$, and is not dependent on induction in any particular case.
A: The OP's claim can be summarized by saying that $n^3$ can be expressed as
the sum of $n$ consecutive odd numbers starting with $n(n+1) - (2n - 1)$ and ending with $n(n+1) - 1$.
This is easy to show by a direct proof, using a result for $n^2$ that
is well-known and easy to prove by induction.
Take the series of odd numbers in reverse order, from largest to
smallest.  Then we are saying that:
\begin{eqnarray}
  n^3 & = & \big[(n^2 + n - 1) + (n^2 + n - 3) + \ldots + (n^2 + n - (2n - 1))\big] \\
      & = & n (n^2 + n) - [1 + 3 + \ldots + 2n - 1]
\end{eqnarray}
This in turn gives:
\begin{equation}
  n^2 = 1 + 3 + \ldots + 2n - 1
\end{equation}
which is an exercise in elementary proof by induction.
A: There are identitity's given below:
For even cube:
$p^3=(p^2-p+1)+\cdots+(p^2-5)+(p^2-3)+(p^2-1)+(p^2+1)+(p^2+3)+(p^2+5)+\cdots+(p^2+p-1)$
For odd cube:
$q^3=(q^2-q+1)+\cdots+(q^2-6)+(q^2-4)+(q^2-2)+(q)^2+(q^2+2)+(q^2+4)+(q^2+6)+\cdots+(q^2+p-1)$
For, $p=8$ we get:
$8^3=(57+59+61+63+65+67+69+71)$
For, $q=9$ we get:
$9^3=(73+75+77+79+81+83+85+87+89)$
A: The proposed identity says that
$n^3
=\sum_{k=0}^{n-1}(n(n-1)+1+2k)
$.
(Figuring out how to write this
is the hard part.)
The right side is
$\sum_{k=0}^{n-1}(n(n-1)+1+2k)
=n(n(n-1)+1)+2\sum_{k=0}^{n-1}k
=n^3-n^2+n+n(n-1)
=n^3
$.
