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 $T_n = 1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1$.

We have $T_1 = 1$

and

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

Thus we have:

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


$\blacksquare$