Cube of 20 is Sum of Sequence of 4 Consecutive Cubes

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Theorem

$20^3 = \displaystyle \sum_{k \mathop = 11}^{14} k^3$

That is:

$20^3 = 11^3 + 12^3 + 13^3 + 14^3$


Proof

\(\displaystyle \sum_{k \mathop = 1}^{14} k^3\) \(=\) \(\displaystyle \paren {\dfrac {14 \paren {14 + 1} } 2}^2\) Sum of Sequence of Cubes
\(\displaystyle \) \(=\) \(\displaystyle 11 \, 025\)
\(\displaystyle \sum_{k \mathop = 1}^{10} k^3\) \(=\) \(\displaystyle \paren {\dfrac {10 \paren {10 + 1} } 2}^2\) Sum of Sequence of Cubes
\(\displaystyle \) \(=\) \(\displaystyle 3025\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{k \mathop = 11}^{14} k^3\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^{14} k^3 - \sum_{k \mathop = 1}^{10} k^3\)
\(\displaystyle \) \(=\) \(\displaystyle 11 \, 025 - 3025\)
\(\displaystyle \) \(=\) \(\displaystyle 8000\)
\(\displaystyle \) \(=\) \(\displaystyle 20^3\)

$\blacksquare$


Sources

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