An algebra problem by Ossama Ismail

$\begin{aligned} n^2 f(n) &= f(1) + f(2) + f(3) + \cdots + f(n) \\ (n^2 -1) f(n) &= f(1) + f(2) + f(3) + \cdots + f(n-1) \\ f(n) &= \dfrac {f(1)+ f(2) + f(3) + \cdots + f(n-1)}{(n^2 -1) } \end{aligned}$

By using simple induction

$\begin{aligned} f(1) &= 255 \\ f(2) = \dfrac {f(1)}{(2^2-1)} = \dfrac {f(1)}{3} &= \dfrac {2 f(1)}{2\cdot 3} \\ f(3) = \dfrac {f(2) + f(2)}{(3^2-1)} = \dfrac {f(1) + \dfrac{f(1)}{3}}{8} = \dfrac {\dfrac{4f(1)}{3}}{8} &= \dfrac{2 f(1)}{3\cdot 4} \\ \cdots \\ f(n) &= \dfrac{2 f(1) }{n(n+1)} \end{aligned}$

then $f(255) = \dfrac{2 \cdot 255}{255 \cdot 256} = \dfrac{2}{256}$

Answer $= \dfrac{1}{f(255)} = 128$

$\sum_{i=1}^n f(i) = n^2 f(n)$

Replacing $n$ by $n+1$ ,

$\sum_{i=1}^{n+1} f(i) = {n+1}^2 f(n+1)$

Subtracting above 2 equations and expanding the square,

$n^2 f(n)=(n^2+2n) f(n+1)$

Thus we have $f(n+1)=\frac{n}{n+2} f(n)$ (You could solve it using reccurence or by trying to find a pattern)

Putting $n \to n+1$

$f(n+2)=\frac{n+1}{n+3} (\frac{n}{n+2} f(n))$

Seeing the pattern:

$f(n+a)= \frac{\prod_{i=n}^{n+a-1} i}{\prod_{j=n+2}^{n+a+1}} f(n)$

\[f(n+a)=\frac{(n)(n+1)}{(n+a)(n+a+1)} f(n)

If n=1 and a=254

\[f(255)=\frac{(1)(2)}{(255)(256)} (255)\]

$f(225)= \frac{1}{128}$

Thus our answer is 128