Sum of Terms of Magic Square

Theorem
The total of all the entries in a magic square of order $n$ is given by:


 * $T_n = \dfrac {n^2 \paren {n^2 + 1} } 2$

Proof
Let $M_n$ denote a magic square of order $n$.

$M_n$ is by definition a square matrix of order $n$ containing the positive integers from $1$ upwards.

Thus there are $n^2$ entries in $M_n$, going from $1$ to $n^2$.

Thus: