 ## 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

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