When a set of natural numbers is under consideration, if we add first consecutive 'n' odd natural numbers(i.e. from 1 ) we get a complete square whose root is 'n' itself.

e.g. first 5 consecutive odd natural numbers are,
1,3,5,7,9 so, 
1+3+5+7+9 = 25.
We get √25 = 5 .
e.g. first 6 consecutive odd natural numbers are,
1,3,5,7,9,11 so,
1+3+5+7+9+11 = 36
We get √36 = 6 .

I observed this while working on an algorithm to identify perfect squares. Has this been observed by any mathematician before?

  • 1
    $\begingroup$ Boy, is this well known. However, your discovering it by yourself is good. Don't worry whether some you discovered is new - it probably is not. That does not matter. $\endgroup$ – marty cohen Aug 24 '15 at 18:52
  • $\begingroup$ Duplicate of this and this. $\endgroup$ – Lucian Aug 25 '15 at 0:03

Yes, it's kind of amazing isn't it! To explain we have: Odd number $2i-1$

$\begin{align} \sum_{i=1}^{n}(2i-1) &=2\sum_{i=1}^{n}i+\sum_{i=1}^{n}1 \\ &=\frac{2(n)(n-1)}{2}+n \\ &=n^{2}-n+n=n^2 \end{align}$


Yes, this fact is widely known. Look, for example, this diagram:

enter image description here

  • $\begingroup$ Nice elaboration.. Thank you! $\endgroup$ – Deepeshkumar Aug 24 '15 at 18:41

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