# 1+2+...+n+(n-1)+...+1 = n^2/Proof 3

## Theorem

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

## 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 our 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 our induction step:

 $\displaystyle$  $\displaystyle 1 + 2 + \cdots + \paren {k + 1} + k + \paren {k - 1} + \cdots + 1$ $\displaystyle$ $=$ $\displaystyle \paren {1 + 2 + \cdots + k + \paren {k - 1} + \cdots + 1} + k + \paren {k + 1}$ $\displaystyle$ $=$ $\displaystyle k^2 + k + \paren {k + 1}$ from induction hypothesis $\displaystyle$ $=$ $\displaystyle k^2 + 2k + 1$ $\displaystyle$ $=$ $\displaystyle \paren {k + 1}^2$

The result follows by induction.

$\blacksquare$