1+2+...+n+(n-1)+...+1 = n^2/Proof 1

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Theorem

$\forall n \in \N: 1 + 2 + \cdots + n + \left({n-1}\right) + \cdots + 1 = n^2$


Proof

\(\displaystyle \) \(\) \(\displaystyle 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1\)
\(\displaystyle \) \(=\) \(\displaystyle 1 + 2 + \cdots + \paren {n - 1} + \paren {n - 1} + \cdots + 1 + n\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {1 + 2 + \cdots + \paren {n - 1} } + n\)
\(\displaystyle \) \(=\) \(\displaystyle 2 \paren {\frac {\paren {n - 1} n} 2} + n\) Closed Form for Triangular Numbers
\(\displaystyle \) \(=\) \(\displaystyle \paren {n - 1} n + n\)
\(\displaystyle \) \(=\) \(\displaystyle n^2 - n + n\)
\(\displaystyle \) \(=\) \(\displaystyle n^2\)

$\blacksquare$