1+2+...+n+(n-1)+...+1 = n^2

Theorem

$\forall n \in \N: 1 + 2 + \cdots + n + \left({n-1}\right) + \cdots + 1 = n^2$

Illustration

Direct Proof 1

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle 1 + 2 + \cdots + \left({n-1}\right) + n + \left({n-1}\right) + \cdots + 1\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 + 2 + \cdots + \left({n-1}\right) + \left({n-1}\right) + \cdots + 1 + n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \left({1 + 2 + \cdots + \left({n-1}\right)}\right) + n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 2 \left({\frac {\left({n-1}\right) n} 2}\right) + n\)	\(\displaystyle \)	Closed Form for Triangular Numbers
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({n-1}\right) n + n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^2 - n + n\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^2\)	\(\displaystyle \)	as desired.

$\blacksquare$

Direct Proof 2

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle 1 + 2 + \cdots + \left({n - 1}\right) + n + \left({n - 1}\right) + \cdots + 1\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({1 + 2 + \cdots + \left({n - 1}\right)}\right) + \left({1 + 2 + \cdots + \left({n - 1}\right) + n}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {\left({n - 1}\right) n} 2 + \frac {n \left({n + 1}\right)} 2\)	\(\displaystyle \)	Closed Form for Triangular Numbers
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {n^2 - n + n^2 + n} 2\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle n^2\)	\(\displaystyle \)	as desired.

$\blacksquare$

Proof by Induction

Proof by induction:

Base case

$n = 1$ holds trivially.

Just to make sure, we try $n = 2$:

$1 + 2 + 1 = 4$

Likewise $n^2 = 2^2 = 4$.

So shown for base case.

Induction Hypothesis

This is our induction hypothesis:

$1 + 2 + \cdots + k + \left({k-1}\right) + \cdots + 1 = k^2$

Now we need to show true for $n=k+1$:

$1 + 2 + \cdots + \left({k + 1}\right) + k + \left({k-1}\right) + \cdots + 1 = \left({k + 1}\right)^2$

Induction Step

This is our induction step:

	\(\displaystyle \)	\(\displaystyle \)	\(\)	\(\displaystyle 1 + 2 + \cdots + \left({k + 1}\right) + k + \left({k-1}\right) + \cdots + 1\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({1 + 2 + \cdots + k + \left({k-1}\right) + \cdots + 1}\right) + k + \left({k + 1}\right)\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle k^2 + k + \left({k + 1}\right)\)	\(\displaystyle \)	from induction hypothesis
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle k^2 + 2k + 1\)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \left({k + 1}\right)^2\)	\(\displaystyle \)

The result follows by induction.

$\blacksquare$