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)?
 A: This diagram may make clear what’s going on:
$$\begin{array}{ccc}
1&\color{red}3&\color{blue}5&\color{orange}7&\color{green}9\\
\color{red}3&\color{red}3&\color{blue}5&\color{orange}7&\color{green}9\\
\color{blue}5&\color{blue}5&\color{blue}5&\color{orange}7&\color{green}9\\
\color{orange}7&\color{orange}7&\color{orange}7&\color{orange}7&\color{green}9\\
\color{green}9&\color{green}9&\color{green}9&\color{green}9&\color{green}9
\end{array}$$
A: 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)\\
&=  \left(\sum_{i=1}^{n+1}(2i-1)\right),
\end{align}
and so the claim holds for all $n$. 
A: 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.
