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
\(\ds \) \(=\) \(\ds \paren {n - 1} n + n\)
\(\ds \) \(=\) \(\ds n^2 - n + n\)
\(\ds \) \(=\) \(\ds n^2\)

$\blacksquare$