## Double An Odd Sum

#### Problem

Consider the following results:

4 (1) | = | 1 + 3 |

4 (1 + 3) | = | 1 + 3 + 5 + 7 |

4 (1 + 3 + 5) | = | 1 + 3 + 5 + 7 + 9 + 11 |

4 (1 + 3 + 5 + 7) | = | 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 |

If $S_n$ represents the sum of the first $n$ odd numbers, prove that $4S_n = S_{2n}$.

#### Solution

We shall prove the result using two informal methods.

__Method 1__

Begin by representing the series 1 + 3 + 5 + 7 + 9 + 11 diagramatically.

O

O O O

O O O O O

O O O O O O O

O O O O O O O O O

O O O O O O O O O O O

In general it can be seen that $S_{2n}$ can be represented by a triangle with $2n$ rows. It is also clear that the top half of triangle, which represents $S_n$, is one quarter of the triangle. Hence $S_{2n} = 4S_n$.

__Method 2__

Consider the following diagram which represents the sum of the first four odd numbers: 1 + 3 + 5 + 7.

O O O O

O O O O

O O O O

O O O O

This leads to the result, $S_n = n^2$.

Therefore $S_{2n} = (2n)^2 = 4n_2 = 4S_n$.

Problem ID: 283 (23 Jul 2006) Difficulty: 2 Star