1+2+...+n+(n-1)+...+1 = n^2/Proof 3
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Theorem
- $\forall n \in \N: 1 + 2 + \cdots + n + \paren {n - 1} + \cdots + 1 = n^2$
Proof
Proof by induction:
Basis for the Induction
$n = 1$ holds trivially.
Just to make sure, we try $n = 2$:
- $1 + 2 + 1 = 4$
Likewise $n^2 = 2^2 = 4$.
So shown for basis for the induction.
Induction Hypothesis
This is the induction hypothesis:
- $1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1 = k^2$
Now we need to show true for $n = k + 1$:
- $1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1 = \paren {k + 1}^2$
Induction Step
This is the induction step:
\(\ds \) | \(\) | \(\ds 1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1} + k + \paren {k + 1}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds k^2 + k + \paren {k + 1}\) | from induction hypothesis | |||||||||||
\(\ds \) | \(=\) | \(\ds k^2 + 2k + 1\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \paren {k + 1}^2\) |
The result follows by induction.
$\blacksquare$