The following question was posed on a national selection competition:
Find the greatest number in the following sequence:
$\frac{1}{503}$, $\frac{4}{524}$, $\frac{9}{581}$, $\frac{16}{692}$...
It took me a long time before I could solve this question, as I could not identify the motif displayed in the above sequence.
Afterwards I realized that the motif is expressed by the function $f(x)=\frac{x^2}{500+3x^3}$.
Once I realized that then, I solved the question as follows:
As $f(x)=\frac{x^2}{500+3x^3}$, then $f'(x)=\frac{2x(500+3x^3)-x^2*9x^2}{(500+3x^3)^2}$
$f'(x)=\frac{1000x-3x^4}{(500+3x^3)^2}$
$f'(x)=\frac{x(1000-3x^3)}{(500+3x^3)^2}$
So all we have to do now is find its maximum. $f'(x)=0$, for $x=\frac{10}{\sqrt[3]{3}}$ and $x=0$, once we analyze the function we find that for $x=\frac{10}{\sqrt[3]{3}}$ we have the maximum value of the function. However since that is a non integral number, we check for $x=6$ and $x=7$, which are the two closest integers to $x=\frac{10}{\sqrt[3]{3}}$.
We have that $f(7)>f(6)$ and hence its maximum is reached for $x=7$ and $f(x)=\frac{49}{1529}$
My question is how do I recognize this motif, as that was my biggest hurdle for this question?
The numerators are clearly the squares.
The $503$ starting the denominators and their slow growth suggests subtracting $500$. Then you see $$ 3, 24, 81, 192, \ldots \ . $$
Perhaps you recognize the triples of the cubes. If not, successive differences soon lead you there.
Granted, this is hindsight.