Question

A function $f(x)$ satisfies $f(1) = 3600$ and $f(1) + f(2) + \dots + f(n) = n^2 f(n)$, for all positive integers $n > 1$. What is the value of $f(9)$?

Accepted Answer

A. Given $f(1) + f(2) + \dots + f(n) = n^2 f(n)$. For $n=2$: $f(1) + f(2) = 4f(2) \implies f(2) = \frac{f(1)}{3}$. By induction, $f(n) = f(1) 	imes \frac{2^2 	imes 3^2 	imes \dots 	imes (n-1)^2}{(2^2-1)(3^2-1)\dots (n^2-1)}$. Simplifying terms yields $f(9) = 3600 	imes \frac{2}{9 	imes 10} = 80$.

Function defined by sum of terms

Explanation

Practise this under exam conditions