Cube of 180 is Sum of Sequence of Consecutive Cubes

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Theorem

$180^3 = \displaystyle \sum_{k \mathop = 6}^{69} k^3$

That is:

$180^3 = 6^3 + 7^3 + \cdots + 67^3 + 68^3 + 69^3$


Proof

\(\displaystyle \sum_{k \mathop = 1}^{69} k^3\) \(=\) \(\displaystyle \paren {\dfrac {69 \paren {69 + 1} } 2}^2\) Sum of Sequence of Cubes
\(\displaystyle \) \(=\) \(\displaystyle 5 \, 832 \, 225\)
\(\displaystyle \sum_{k \mathop = 1}^5 k^3\) \(=\) \(\displaystyle \paren {\dfrac {5 \paren {5 + 1} } 2}^2\) Sum of Sequence of Cubes
\(\displaystyle \) \(=\) \(\displaystyle 225\)
\(\displaystyle \leadsto \ \ \) \(\displaystyle \sum_{k \mathop = 6}^{69} k^3\) \(=\) \(\displaystyle \sum_{k \mathop = 1}^{69} k^3 - \sum_{k \mathop = 1}^5 k^3\)
\(\displaystyle \) \(=\) \(\displaystyle 5 \, 832 \, 225 - 225\)
\(\displaystyle \) \(=\) \(\displaystyle 5 \, 832 \, 000\)
\(\displaystyle \) \(=\) \(\displaystyle 180^3\)

$\blacksquare$


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