Sum of Sequence of Cubes divides 3 times Sum of Sequence of Fifth Powers
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Theorem
- $\ds \sum_{i \mathop = 1}^n i^3 \divides 3 \sum_{i \mathop = 1}^n i^5$
where $\divides$ denotes divisibility.
Proof
\(\ds 3 \sum_{i \mathop = 1}^n i^5\) | \(=\) | \(\ds {T_n}^2 \paren {4 T_n - 1}\) | Sum of Sequence of Fifth Powers | |||||||||||
\(\ds \) | \(=\) | \(\ds k {T_n}^2\) | where $k = 4 T_n - 1$ | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds {T_n}^2\) | \(\divides\) | \(\ds 3 \sum_{i \mathop = 1}^n i^5\) | Definition of Divisor of Integer | ||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{i \mathop = 1}^n i^3\) | \(\divides\) | \(\ds 3 \sum_{i \mathop = 1}^n i^5\) | Square of Triangular Number equals Sum of Sequence of Cubes |
$\blacksquare$
Sources
- 1970: Wacław Sierpiński: 250 Problems in Elementary Number Theory: No. $9$
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$