2860.41 – Fraction Pattern

We use the fact that the sum of the first $n$ odd numbers is $n^2$. For instance, $1 + 3 + 5 + 7 = 16 = 4^2$.

We also use the fact (this is tricky) that, say, $7 + 9 + 11$ can be thought of as $1 + 3 + 5 + 7 + 9 + 11 - (1 + 3 + 5)$, i.e. $(6^2-3^2)$.

Thus the denominator in each of these fractions is the sum of the first $2n$ odd numbers minus the sum of the first $n$ odd numbers:

$$\begin{align*}\dfrac{1+3}{5+7} &= \dfrac{1+3}{1+3+5+7-(1+3)} \\[1em]&= \dfrac{2^2}{4^2-2^2} \\[1em]&= \dfrac{4}{16-4} \\[1em]&= \dfrac{4}{12} \\[1em]&= \dfrac{1}{3}\end{align*}$$

Aha! And:

$$\begin{align*}\dfrac{1+3+5+7}{9+11+13+15} &= \dfrac{1+3+5+7}{1+3+5+\cdots+13+15-(1+3+5+7)} \\[1em]&= \dfrac{4^2}{8^2-4^2} \\[1em]&= \dfrac{16}{64-16} \\[1em]&= \dfrac{16}{48} \\[1em]&= \dfrac{1}{3}\end{align*}$$

YES!

We are heading toward the formula $\dfrac{n^2}{(2n)^2-n^2} = \dfrac{n^2}{4n^2-n^2} = \dfrac{n^2}{3n^2} = \dfrac{1}{3}$.