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Question 17
A function $$f (x)$$ satisfies $$f(1) = 3600$$, and $$f (1) + f(2) + ... + f(n) =n^2f(n)$$, for all positive integers $$n > 1$$. What is the value of $$f (9)$$ ?
Solution
According to given conditions we get f(2)=f(1)/3 , then f(3)=f(1)/6, then  f(4)=f(1)/10 , then f(5)=f(1)/15 .
We can see the pattern here that the denominator goes on increasing from 3,3+3,6+4,10+5,15+6,.. so for the f(9) the denominator will be same as 15+6+7+8+9=45 .
So f(9)=3600/45 = 80
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