Cube of 20 is Sum of Sequence of 4 Consecutive Cubes
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Theorem
- $20^3 = \ds \sum_{k \mathop = 11}^{14} k^3$
That is:
- $20^3 = 11^3 + 12^3 + 13^3 + 14^3$
Proof
\(\ds \sum_{k \mathop = 1}^{14} k^3\) | \(=\) | \(\ds \paren {\dfrac {14 \paren {14 + 1} } 2}^2\) | Sum of Sequence of Cubes | |||||||||||
\(\ds \) | \(=\) | \(\ds 11 \, 025\) | ||||||||||||
\(\ds \sum_{k \mathop = 1}^{10} k^3\) | \(=\) | \(\ds \paren {\dfrac {10 \paren {10 + 1} } 2}^2\) | Sum of Sequence of Cubes | |||||||||||
\(\ds \) | \(=\) | \(\ds 3025\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds \sum_{k \mathop = 11}^{14} k^3\) | \(=\) | \(\ds \sum_{k \mathop = 1}^{14} k^3 - \sum_{k \mathop = 1}^{10} k^3\) | |||||||||||
\(\ds \) | \(=\) | \(\ds 11 \, 025 - 3025\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 8000\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds 20^3\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $8000$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $8000$