A property regarding complete/perfect squares.

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 .
or
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?

• 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. – marty cohen Aug 24 '15 at 18:52
• Duplicate of this and this. – 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}