Find a closed form for the equations $1^3 = 1$, $2^3 = 3 + 5$, $3^3 = 7 + 9 + 11$

Question

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...

Answer 1

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, $$\sum_{r=0}^{m-1}\{(m^2-m+1)+2r\}=(m^2-m+1)\sum_{r=0}^{m-1}1+2\sum_{r=0}^{m-1}r$$

$$=m(m^2-m+1)+2\cdot\frac{m(m-1)}2=m^3$$

Answer 2

$$n^3=\sum_{k=0}^{n-1}(n^2-(n-1)+2k)$$

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.

Answer 3

$$\begin{align} 1^3&=1\\ 2^3&=3+5\\ 3^3&=7+9+11\\ 4^3&=13+15+17+19\\ 5^3&=21+23+25+27+29\\ \vdots &= \vdots\\ n^3&=[(n^2-n+1)]+[(n^2-n+1)+2]+[(n^2-n+1)+4]+\cdots+[(n^2-n+1)+2(n-1)]\\ &=\sum_{r=1}^n(n^2-n+1)+2(r-1)\qquad \blacksquare \end{align}$$

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:

$$\begin{align} \text{RHS}&=\sum_{r=1}^n[\color{blue}{(n^2-n+1)}+2(r\color{blue}{-1})]\\ &=n(n^2-n-1)+2\sum_{r=1}^n r\\ &=n(n^2-n-1)+2\cdot \frac {n(n+1)}2\\ &=n^3=\text{LHS}\qquad \blacksquare \end{align}$$