Cube of 180 is Sum of Sequence of Consecutive Cubes

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 \ \$ $\displaystyle \sum_{k \mathop = 6}^{69} k^3$ $=$ $\displaystyle \sum_{k \mathop = 1}^{69} k^3 - \sum_{k \mathop = 1}^5 k^3$ $\displaystyle$ $=$ $\displaystyle 5 \, 832 \, 225 - 225$ $\displaystyle$ $=$ $\displaystyle 5 \, 832 \, 000$ $\displaystyle$ $=$ $\displaystyle 180^3$

$\blacksquare$