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

## Theorem

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

## Direct Proof 1

 $\displaystyle$  $\displaystyle 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1$ $\displaystyle$ $=$ $\displaystyle 1 + 2 + \cdots + \paren {n - 1} + \paren {n - 1} + \cdots + 1 + n$ $\displaystyle$ $=$ $\displaystyle 2 \paren {1 + 2 + \cdots + \paren {n - 1} } + n$ $\displaystyle$ $=$ $\displaystyle 2 \paren {\frac {\paren {n - 1} n} 2} + n$ Closed Form for Triangular Numbers $\displaystyle$ $=$ $\displaystyle \paren {n - 1} n + n$ $\displaystyle$ $=$ $\displaystyle n^2 - n + n$ $\displaystyle$ $=$ $\displaystyle n^2$

$\blacksquare$

## Direct Proof 2

 $\displaystyle$  $\displaystyle 1 + 2 + \cdots + \paren {n - 1} + n + \paren {n - 1} + \cdots + 1$ $\displaystyle$ $=$ $\displaystyle \paren {1 + 2 + \cdots + \paren {n - 1} } + \paren {1 + 2 + \cdots + \paren {n - 1} + n}$ $\displaystyle$ $=$ $\displaystyle \frac {\paren {n - 1} n} 2 + \frac {n \paren {n + 1} } 2$ Closed Form for Triangular Numbers $\displaystyle$ $=$ $\displaystyle \frac {n^2 - n + n^2 + n} 2$ $\displaystyle$ $=$ $\displaystyle n^2$

$\blacksquare$

## Proof by Induction

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.

$\blacksquare$

## Proof 4

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

$\blacksquare$