A property regarding complete/perfect squares.

Question

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

score 3 · Answer 1 · answered Aug 24 '15 at 18:26

3

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

answered Aug 24 '15 at 18:26

ajotatxe

65,084

Nice elaboration.. Thank you! – Deepeshkumar Aug 24 '15 at 18:41

score 2 · Accepted Answer · edited Aug 24 '15 at 18:41

2

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

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

edited Aug 24 '15 at 18:41

vonbrand

27,812

answered Aug 24 '15 at 18:35

Socre

559

A property regarding complete/perfect squares.

2 Answers2