1+2+...+n+(n-1)+...+1 = n^2/Proof 4

Theorem

 * $\forall n \in \N: 1 + 2 + \cdots + n + \left({n-1}\right) + \cdots + 1 = n^2$

Proof
Let $T_n = 1 + 2 + \cdots + n + \left({n-1}\right) + \cdots + 1$.

We have $T_n = 1$

and

The result follows from Odd Number Theorem and Principle of Recursive Definition.