# Sum of Odd Numbers make Squares

Look at this:

1
(+3)
4
(+5)
9
(+7)
16
(+9)
25
(+11)
36
(+13)
49

And so forth, you get the idea.

Why do they make up this pattern? And is there any special name for this type of sequence (adding up every odd number/even number/nth number)?

• Take any checkered sheet of paper and fill out the squares in the following way: Fill out any square, then fill out the adjacent L-shape of three squares, then fill out the adjacent L-shape of five squares … Ok, like indicated by Brian M. Scott. Oct 2, 2015 at 23:55
• Since nobody's mentioned it, those are all examples of arithmetic series, where each term differs from the previous by a constant amount. Oct 3, 2015 at 0:04
• I am now faced with the dilemma: Accept the easy-to-understand answer, or the hardcore math proof. Oct 3, 2015 at 0:21

$$\begin{array}{ccc} 1&\color{red}3&\color{blue}5&\color{orange}7&\color{green}9\\ \color{red}3&\color{red}3&\color{blue}5&\color{orange}7&\color{green}9\\ \color{blue}5&\color{blue}5&\color{blue}5&\color{orange}7&\color{green}9\\ \color{orange}7&\color{orange}7&\color{orange}7&\color{orange}7&\color{green}9\\ \color{green}9&\color{green}9&\color{green}9&\color{green}9&\color{green}9 \end{array}$$
The pattern here is $$n^2 = \sum_{i=1}^{n}(2i-1)$$ for positive integers $$n$$. To rigorously prove this, we can use mathematical induction. It is clear that $$1^2 = \sum_{i=1}^1 (2i-1) = 1$$. Now assuming that $$n^2 =\sum_{i=1}^n (2i-1)$$ for some $$n\geqslant 1$$, we have \begin{align} (n+1)^2 &= n^2 + 2n + 1\\ &= \left(\sum_{i=1}^n(2i-1)\right) + (2n+1)\\ &= \left(\sum_{i=1}^n(2i-1)\right) + (2(n+1)-1)\\ &= \sum_{i=1}^{n+1}(2i-1), \end{align} and so the claim holds for all $$n$$.