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
Proof by induction:

For all $$n \in \N^*$$, let $$P \left({n}\right)$$ be the proposition $$n^2 = \sum_{j=1}^n \left({2j-1}\right)$$.

Basis for the Induction

 * $$P(1)$$ is true, as this just says $$1^2 = 1$$.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 1$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

So this is our induction hypothesis:

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

Then we need to show:

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

Induction Step
This is our induction step:

$$ $$ $$ $$

So $$P \left({k}\right) \implies P \left({k+1}\right)$$ and the result follows by the Principle of Mathematical Induction.

Therefore $$\forall n \in \N: n^2 = \sum_{j=1}^n \left({2j-1}\right)$$.

Proof of Corollary
Follows directly.

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


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