Square of Triangular Number equals Sum of Sequence of Cubes/Proof 1
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Theorem
- $\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
where $T_n$ denotes the $n$th triangular number.
Proof
| \(\ds \sum_{i \mathop = 1}^n i^3\) | \(=\) | \(\ds \frac {n^2 \paren {n + 1}^2} 4\) | Sum of Sequence of Cubes | |||||||||||
| \(\ds \) | \(=\) | \(\ds \paren {\frac {n \paren {n + 1} } 2}^2\) | ||||||||||||
| \(\ds \) | \(=\) | \(\ds {T_n}^2\) | Closed Form for Triangular Numbers |
$\blacksquare$