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