Recurrence Formula for Square Numbers

Theorem
Let $$S_n$$ denote the $$n$$th square number, i.e. let $$S_n = n^2$$.

Then $$n^2 = \sum_{j=1}^n \left({2j-1}\right)$$.

Corollary
Then $$S_n = S_{n-1} + 2n - 1$$.

Proof of Main Theorem
This is demonstrated in Sum of Sequence of Odd Numbers.

Proof of Corollary
Follows directly.

Comment
What this shows is that every square number is the sum of a series of consecutive odd integers:


 * $$n^2 = 1 + 3 + 5 + \cdots + \left({2n-1}\right)$$