
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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