Cube Number as Difference between Squares of Triangular Numbers
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Theorem
Let $n \in \Z_{> 0}$ be a positive integer.
Then:
- $n^3 = {T_n}^2 - {T_{n - 1} }^2$
where $T_n$ denotes the $n$th triangular number.
Proof
\(\displaystyle {T_n}^2 - {T_{n - 1} }^2\) | \(=\) | \(\displaystyle \paren {\frac {k \paren {k + 1} } 2}^2 - \paren {\frac {\paren {k - 1} k} 2}^2\) | Closed Form for Triangular Numbers | ||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\paren {k^2 + k}^2 - \paren {k^2 - k}^2} 4\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\paren {k^4 + 2 k^3 + k^2} - \paren {k^4 - 2 k^3 + k^2} } 4\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle \frac {4 k^3} 4\) | |||||||||||
\(\displaystyle \) | \(=\) | \(\displaystyle k^3\) |
$\blacksquare$
Sources
- 1986: David Wells: Curious and Interesting Numbers ... (previous) ... (next): $15$
- 1997: David Wells: Curious and Interesting Numbers (2nd ed.) ... (previous) ... (next): $15$