1+2+...+n+(n-1)+...+1 = n^2/Proof 4
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Theorem
- $\forall n \in \N: 1 + 2 + \cdots + n + \left({n-1}\right) + \cdots + 1 = n^2$
Proof
Let $T_n = 1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1$.
We have $T_1 = 1$
and
| \(\displaystyle T_n - T_{n-1}\) | \(=\) | \(\displaystyle \left({1 + 2 + \cdots + n + \left({n - 1}\right) + \cdots + 1 }\right)\) | Definition of $T_n$ | ||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle - \, \) | \(\displaystyle \left({1 + 2 + \cdots + \left({n - 1}\right) + \left({n - 2}\right) + \cdots + 1}\right)\) | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({\left({1 + 2 + \cdots + n}\right) - \left({1 + 2 + \cdots + \left({n - 1}\right)}\right)}\right)\) | Integer Addition is Associative | ||||||||||
| \(\displaystyle \) | \(\) | \(\, \displaystyle + \, \) | \(\displaystyle \left({\left({\left({n - 1}\right) + \left({n - 2}\right) + \cdots + 1}\right) - \left({\left({n - 2}\right) + \left({n - 3}\right) + \cdots + 1}\right)}\right)\) | Integer Addition is Commutative | |||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle n + \left({n - 1}\right)\) | simplifying | ||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle 2 n - 1\) |
Thus we have:
| \(\displaystyle T_n\) | \(=\) | \(\displaystyle \left({T_n - T_{n-1} }\right) + \left({T_{n-1} - T_{n-2} }\right) + \cdots + \left({T_2 - T_1}\right) + T_1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \left({2 n - 1}\right) + \left({2 \left({n - 1}\right) - 1}\right) + \cdots + \left({2 \times 2 - 1}\right) + 1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle \sum_{k \mathop = 1}^n 2 k - 1\) | |||||||||||
| \(\displaystyle \) | \(=\) | \(\displaystyle n^2\) | Odd Number Theorem |
$\blacksquare$
