# Sum of Sequence of Cubes divides 3 times Sum of Sequence of Fifth Powers

## Theorem

$\displaystyle \sum_{i \mathop = 1}^n i^3 \divides 3 \sum_{i \mathop = 1}^n i^5$

where $\divides$ denotes divisibility.

## Proof

 $\displaystyle 3 \sum_{i \mathop = 1}^n i^5$ $=$ $\displaystyle {T_n}^2 \paren {4 T_n - 1}$ Sum of Sequence of Fifth Powers $\displaystyle$ $=$ $\displaystyle k {T_n}^2$ where $k = 4 T_n - 1$ $\displaystyle \leadsto \ \$ $\displaystyle {T_n}^2$ $\divides$ $\displaystyle 3 \sum_{i \mathop = 1}^n i^5$ Definition of Divisor of Integer $\displaystyle \leadsto \ \$ $\displaystyle \sum_{i \mathop = 1}^n i^3$ $\divides$ $\displaystyle 3 \sum_{i \mathop = 1}^n i^5$ Square of Triangular Number equals Sum of Sequence of Cubes

$\blacksquare$