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 |
$\blacksquare$