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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. $q = 2k + 1.$

Martin Sleziak
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base 10
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    Well, the base case is half of the work: and you have just written it down. So you are left with the inductive step. – Crostul May 28 '18 at 11:31
  • Not really by examining the base case for example : $\sum_{i=1}^{j =n} (2^{n-1} + q_{j}) = n^3 $ when n =1 $ 2^{n-1} + q_{1} = 1^3 , \ thus \ q_{1} \ is \ even \ which \ contradicts \ the \ statement \ i \ made \ that \ q \ is \ odd$ – base 10 May 28 '18 at 16:54

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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:

Martin Sleziak
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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 $$\sum_{m=\frac{n^2-n+1}2-1}^{\frac{n^2+n-1}2-1}(2m+1)=?$$

lab bhattacharjee
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