A 1996 BMO Problem Adaptation

image containing math problem statement
This is an adaptation of a problem which appeared in BMO in the year 1996 This was a relatively harder problem for that exam. Try this problem live now, or view the solution here...
Let us first try to get an expression for the $n\xasuper{th}$ term as a function of the previous terms.

$f(1) + f(2) + f(3) + ... + f(n-1) + f(n) = n^2 f(n)$

$\Rightarrow f(1) + f(2) + f(3) + ... + f(n-1) = n^2 f(n) - f(n)$

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$\Rightarrow f(n) = \dfrac{1}{n^2 - 1} \times [f(1) + f(2) + f(3) + ... + f(n-1)]$

We can now find the first few terms of this series from the given information:
  
$f(2) = \dfrac{1}{2^2-1} f(1)$

$f(3) = \dfrac{1}{3^2-1} \left[f(1) + f(2) \right]$

$= \dfrac{1}{3^2-1} \left[f(1) + \dfrac{1}{2^2-1} f(1) \right]$

$= \dfrac{1}{3^2-1} \left[1 + \dfrac{1}{2^2-1} \right] f(1)$

$= \dfrac{1}{3^2-1} \left[\dfrac{2^2}{2^2-1} \right] f(1)$

$= \dfrac{1}{2^2-1} \dfrac{2^2}{3^2-1} f(1)$

Similarly, we get:
$f(4) = \left( \dfrac{1}{2^2-1} \right) \left(\dfrac{2^2}{3^2-1} \right) \left( \dfrac{3^2}{4^2-1} \right)  f(1)$

$...$

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For $f(n)$ we get:
$f(n) = \left( \dfrac{1^2}{2^2-1} \right) \left(\dfrac{2^2}{3^2-1} \right) \left( \dfrac{3^2}{4^2-1} \right)...\left( \dfrac{(n-1)^2}{n^2-1} \right)  f(1)$

$= \left( \dfrac{(2-1)^2}{2^2-1} \right) \left(\dfrac{(3-1)^2}{3^2-1} \right) \left( \dfrac{(4-1)^2}{4^2-1} \right)...\left( \dfrac{(n-1)^2}{n^2-1} \right) f(1)$

$= \left( \dfrac{(2-1)}{2 + 1} \right) \left(\dfrac{(3-1)}{3 + 1} \right) \left( \dfrac{(4-1)}{4 + 1} \right)...\left( \dfrac{(n-1)}{n + 1} \right) f(1)$

$= \left( \dfrac{1}{\cancel{3}} \right) \left(\dfrac{2}{\cancel{4}} \right) \left( \dfrac{\cancel{3}}{\cancel{5}} \right)...\left( \dfrac{\cancel{n-1}}{n + 1} \right) f(1)$

Observe that each denominator, except for the last two terms, will cancel out with the numerator of its next to next term.
Cancelling out, and simplifying the above terms, we get:

$= \dfrac{1 \times 2}{n(n+1)} f(1)$

$\therefore f(2024) = \dfrac{1 \times 2}{ 2024 \times 2025} \times 2024$

$= \dfrac{2}{ 2025} $

Therefore, $m = 2$, $n = 2025$ and $(m + n) = 2027$