Sum of Consecutive Odd Numbers

Problem

Sum of consecutive odd numbers forming perfect squares

Observe the following pattern:

$$\begin{align} &1 = 1 \\ &1 + 3 = 4 \\ &1 + 3 + 5 = 9 \\ &1 + 3 + 5 + 7 = 16 \\ &1 + 3 + 5 + 6 + 7 + 9 = 25. \end{align}$$

In each equation, the left hand side is the sum of consecutive odd numbers; the right hand side are perfect squares. Prove that the sum of consecutive odd numbers (beginning with 1) is a perfect square.

Solution

The sum of consecutive odd numbers is an arithmetic series. There are two important equations to use:

$$\begin{align} &(1) \; t_n = t_1 + (n-1) d \\[5 pt] &(2) \; S_n = \frac{n}{2}(t_1 + t_n) \end{align}$$

The first term is $t_1 = 1$ and the common difference is $d=2$. This means the final term is $t_n = 1 + (n-1)2 = 2n -1$. Therefore,

$$\begin{align} 1 + 3 + 5 + \cdots + 2n - 1 &= \frac{n}{2}\left(1 + 2n-1 \right) \\[5 pt] 1 + 3 + 5 + \cdots + 2n - 1 &= \frac{n}{2}\times 2n \\[5 pt] 1 + 3 + 5 + ... + 2n - 1 &= n^2. \end{align}$$