1+2+...+n+(n-1)+...+1 = n^2/Proof 2
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Theorem
- $\forall n \in \N: 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 = n^2$
Proof
\(\ds \) | \(\) | \(\ds 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 + 2 + \cdots + \paren {n - 1} } + \paren {1 + 2 + \cdots + \paren {n - 1} + n}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2\) | Closed Form for Triangular Numbers | |||||||||||
\(\ds \) | \(=\) | \(\ds \frac {n^2 - n + n^2 + n} 2\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds n^2\) |
$\blacksquare$