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In the previous post, The Joy of Homogeneity, I followed Gary Davis in establishing a statement observe by Ben Vitale. Ben's observation had to do with fractions in which both the numerator and denominator were sums of consecutive odd numbers. So that, for example,

$$\frac{1+3}{5+7}=\frac{1+3+5}{7+9+11}=\frac{1+3+5+7}{9+11+13+15}= \ldots = \frac{1}{3},$$

and, more generally,

$$\frac{1+2+\ldots +(2n-1)}{(2n+1)+(2n+3)+\ldots +(4n-1)}=\frac{n^2}{3n^2}=\frac{1}{3}.$$

Allen Pinkall left a commenet on the original page with an observation on a regularity in a somewhat modified fraction:

$$\frac{1+2+\ldots +(2n-1)}{(2n-1)+(2n+3)+\ldots +(4n-3)},$$

where the denominator starts with the last term of the numerator and not with the next one. The new observation concerns the sum of the numerator and the denominator of the reduced fraction. Let

$$A = 1+2+\ldots +(2n-1)=n^2$$

and

$$B = (2n-1)+\ldots +(4n-3)=(2n-1)^{2}-(n-1)^{2}=3n^{2}-2n.$$

Now then $\frac{A}{B}=\frac{n^{2}}{n(3n-2)}=\frac{n}{3n-2}=\frac{a}{b}$, where $a = n$ and $b=3n-2$. What we note is that the sum $a+b$ is twice the last term in the numerator (as well as the first term in the numerator.) This is always true. Furthermore, the fraction $\frac{a}{b}$ is irreduceable unless $n$ is even. For $n$ even, the fraction $\frac{a}{b}$ can be further reduced by just a factor of $2$:

$$\frac{n}{3n-2}=\frac{n/2}{3(n/2)-1}$$

such that the sum of the numerator and the denominator of the latter is $(n/2)+3(n/2)-1=2n-1$, exactly the last term of the sum in the numerator (as well as the first term in the numerator.)