Sum of Odd Numbers make Squares

Question

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)?

Answer 1

This diagram may make clear what’s going on:

$$\begin{array}{ccc} 1&\color{red}3&\color{blue}5&\color{orange}7&\color{green}9\\ \color{red}3&\color{red}3&\color{blue}5&\color{orange}7&\color{green}9\\ \color{blue}5&\color{blue}5&\color{blue}5&\color{orange}7&\color{green}9\\ \color{orange}7&\color{orange}7&\color{orange}7&\color{orange}7&\color{green}9\\ \color{green}9&\color{green}9&\color{green}9&\color{green}9&\color{green}9 \end{array}$$

Answer 2

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 $n$.

Answer 3

Another way to see it: Write it as 1+ 3+ 5+ ...+ (2n- 3)+ (2n- 1)+ (2n+ 1)= S and reverse the order (2n+ 1)+ (2n- 1)+ (2n- 3)+...+ 5+ 3+ 1= S. Now add VERTICALLY.

We have 1+ (2n+ 1)= 2n+ 2, 3+ (2n- 1)= 2n+ 2, 5+ (2n-3)= 2n+ 2, ...,

That is, every pair sums to 2n+ 2 and there are n+1 pairs (1= 2(0)+ 1 so the numbering starts at 0, not 1). So the two equations sum to (2n+ 2)(n+ 1)= 2S. Dividing both sides by 2, (n+ 1)(n+ 1)= (n+ 1)^2= S.