Cube of 180 is Sum of Sequence of Consecutive Cubes
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Theorem
- $180^3 = \ds \sum_{k \mathop = 6}^{69} k^3$
That is:
- $180^3 = 6^3 + 7^3 + \cdots + 67^3 + 68^3 + 69^3$
Proof
\(\ds \sum_{k \mathop = 1}^{69} k^3\) | \(=\) | \(\ds \paren {\dfrac {69 \paren {69 + 1} } 2}^2\) | Sum of Sequence of Cubes | |||||||||||
\(\ds \) | \(=\) | \(\ds 5 \, 832 \, 225\) | ||||||||||||
\(\ds \sum_{k \mathop = 1}^5 k^3\) | \(=\) | \(\ds \paren {\dfrac {5 \paren {5 + 1} } 2}^2\) | Sum of Sequence of Cubes | |||||||||||
\(\ds \) | \(=\) | \(\ds 225\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{k \mathop = 6}^{69} k^3\) | \(=\) | \(\ds \sum_{k \mathop = 1}^{69} k^3 - \sum_{k \mathop = 1}^5 k^3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 5 \, 832 \, 225 - 225\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 5 \, 832 \, 000\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 180^3\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $180$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $180$