1+2+...+n+(n-1)+...+1 = n^2/Proof 4
Jump to navigation
Jump to search
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 |
$\blacksquare$