Prove the theorem of Nicomachus by induction Prove the theorem of Nicomachus(AD.100) by induction:
$$ 
1^3 = 1,\  2^3 = 3+ 5,\  3^3 = 7 + 9 + 11,\  4^3 = 13 + 15 + 17 + 19,\  ...
$$
My approach: from looking at the above pattern you can tell there is something of the following sort:
 $$ 2^{n-1} + q = n ^3,$$
where $q$ is odd s.t. $q = 2k + 1.$
 A: Notice that by adding your equations you get
\begin{align*}
1^3&=1\\
1^3+2^3&=1+3+5\\
1^3+2^3+3^3&=1+3+5+7+9+11\\
1^3+2^3+3^3+4^3&=1+3+5+7+9+11+13+15+17+19
\end{align*}
And notice also that these equations imply the equations you want. (You get them simply by subtracting the two consecutive equations.)
So you actually want to prove that sum of the first $n$ cubes is the same as the sume of the first $1+2+\dots+n=\frac{n(n+1)}2$ odd numbers.
$$1^3+2^3+\dots+n^3 = \sum_{k=1}^{n(n+1)/2} (2k-1)$$
If you also know that the sum of the first $n$ odd numbers gives a square, this is the same as
$$1^3+2^3+\dots+n^3 = \left(\frac{n(n+1)}2\right)^2$$
See also:


*

*Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction

*Direct Proof that $1 + 3 + 5 + \cdots+ (2n - 1) = n\cdot n$
A: If induction is not mandatory,
let the $r$th term$(T_r)$ of $1,3,7,13$  be $$a(r-1)(r-2)+b(r-1)+c$$
$r=1\implies 1=c$
$r=2\implies3=b+c\iff b=2$
$r=3\implies7=2a+2b+c\iff a=1$
$$T_r=(r-1)(r-2)+2(r-1)+1=r^2-r+1$$
Sum of $n$ term $$\sum_{m=\frac{n^2-n+1}2-1}^{\frac{n^2+n-1}2-1}(2m+1)=?$$
