Sum of Terms of Magic Cube

Theorem

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

$T_n = \dfrac {n^3 \paren {n^3 + 1} } 2$

Proof

Let $M_n$ denote a magic cube of order $n$.

$M_n$ is by definition an arrangement of the first $n^3$ (strictly) positive integers into an $n \times n \times n$ cubic array containing the positive integers from $1$ upwards.

Thus there are $n^3$ entries in $M_n$, going from $1$ to $n^3$.

Thus:

\(\ds T_n\)

\(=\)

\(\ds \sum_{k \mathop = 1}^{n^3} k\)

\(\ds \)

\(=\)

\(\ds \frac {n^3 \paren {n^3 + 1} } 2\)

Closed Form for Triangular Numbers

$\blacksquare$

Sequence

The sequence of the sum totals of all the entries in magic cubes of order $n$ begins:

$1, \paren {36,} \, 378, 2080, 7875, 23 \, 436, 58 \, 996, 131 \, 328, \ldots$

However, note that while $36 = \dfrac {2^3 \paren {2^3 + 1} } 2$, a magic cube of order $2$ does not actually exist.