# Cube as Sum of Sequence of Centered Hexagonal Numbers

## Theorem

$C_n = \displaystyle \sum_{i \mathop = 1}^n H_i$

where:

$C_n$ denotes the $n$th cube number
$H_i$ denotes the $i$th centered hexagonal number.

## Proof

$H_n = 3 n \paren {n - 1} + 1$

Hence:

 $\displaystyle \sum_{i \mathop = 1}^n H_i$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \paren {3 i \paren {i - 1} + 1}$ $\displaystyle$ $=$ $\displaystyle \sum_{i \mathop = 1}^n \paren {3 i^2 - 3 i + 1}$ $\displaystyle$ $=$ $\displaystyle 3 \sum_{i \mathop = 1}^n i^2 - 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1$ $\displaystyle$ $=$ $\displaystyle 3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \sum_{i \mathop = 1}^n i + \sum_{i \mathop = 1}^n 1$ Sum of Sequence of Squares $\displaystyle$ $=$ $\displaystyle 3 \frac {n \paren {n + 1} \paren {2 n + 1} } 6 - 3 \frac {n \paren {n + 1} } 2 + \sum_{i \mathop = 1}^n 1$ Closed Form for Triangular Numbers $\displaystyle$ $=$ $\displaystyle \frac {n \paren {\paren {n + 1} \paren {2 n + 1} - 3 \paren {n + 1} + 2} } 2$ simplification $\displaystyle$ $=$ $\displaystyle \frac {n \paren {2 n^2 + 3 n + 1 - 3 n - 3 + 2} } 2$ multiplying out $\displaystyle$ $=$ $\displaystyle \frac {n \paren {2 n^2} } 2$ simplification $\displaystyle$ $=$ $\displaystyle n^3$ simplification

$\blacksquare$

## Examples

 $\displaystyle 1^3$ $=$ $\displaystyle 1$ $\displaystyle$ $=$ $\displaystyle 3 \times 1 \paren {1 - 1} + 1$

 $\displaystyle 2^3$ $=$ $\displaystyle 8$ $\displaystyle$ $=$ $\displaystyle 1 + 7$ $\displaystyle$ $=$ $\displaystyle \paren {3 \times 1 \paren {1 - 1} + 1}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {3 \times 2 \paren {2 - 1} + 1}$

 $\displaystyle 3^3$ $=$ $\displaystyle 27$ $\displaystyle$ $=$ $\displaystyle 1 + 7 + 19$ $\displaystyle$ $=$ $\displaystyle \paren {3 \times 1 \paren {1 - 1} + 1}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {3 \times 2 \paren {2 - 1} + 1}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {3 \times 3 \paren {3 - 1} + 1}$

 $\displaystyle 4^3$ $=$ $\displaystyle 64$ $\displaystyle$ $=$ $\displaystyle 1 + 7 + 19 + 37$ $\displaystyle$ $=$ $\displaystyle \paren {3 \times 1 \paren {1 - 1} + 1}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {3 \times 2 \paren {2 - 1} + 1}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {3 \times 3 \paren {3 - 1} + 1}$ $\displaystyle$  $\, \displaystyle + \,$ $\displaystyle \paren {3 \times 4 \paren {4 - 1} + 1}$