1+2+...+n+(n-1)+...+1 = n^2/Proof 4

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Theorem

$\forall n \in \N: 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 = n^2$


Proof

Let $T_n = 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1$.

We have $T_1 = 1$

and

\(\ds T_n - T_{n - 1}\) \(=\) \(\ds \paren {1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 }\) Definition of $T_n$
\(\ds \) \(\) \(\, \ds - \, \) \(\ds \paren {1 + 2 + \cdots + \paren {n - 1} + \paren {n - 2} + \cdots + 1}\)
\(\ds \) \(=\) \(\ds \paren {\paren {1 + 2 + \cdots + n} - \paren {1 + 2 + \cdots + \paren {n - 1} } }\) Integer Addition is Associative
\(\ds \) \(\) \(\, \ds + \, \) \(\ds \paren {\paren {\paren {n - 1} + \paren {n - 2} + \cdots + 1} - \paren {\paren {n - 2} + \paren {n - 3} + \cdots + 1} }\) Integer Addition is Commutative
\(\ds \) \(=\) \(\ds n + \paren {n - 1}\) simplifying
\(\ds \) \(=\) \(\ds 2 n - 1\)

Thus we have:

\(\ds T_n\) \(=\) \(\ds \paren {T_n - T_{n - 1} } + \paren {T_{n - 1} - T_{n - 2} } + \cdots + \paren {T_2 - T_1} + T_1\)
\(\ds \) \(=\) \(\ds \paren {2 n - 1} + \paren {2 \paren {n - 1} - 1} + \cdots + \paren {2 \times 2 - 1} + 1\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 1}^n 2 k - 1\)
\(\ds \) \(=\) \(\ds n^2\) Odd Number Theorem

$\blacksquare$