Olympiad question on functions

Question

OK, I've tried solving the question, and here's what I've been able to do so far:

But I'm lost here.

What do I do now?

Answer 1

Hint : what is $n^2f(n)-(n-1)^2f(n-1)$ ?

In a more general case, when you have this kind of summations, always try to simplify terms. It can help.

Answer 2

$$f(1)+\ldots+f(n)=n^2f(n)$$ $$f(1)+\ldots+f(n-1)=(n-1)^2f(n-1)$$

Subtract both equations and we get

$$f(n)=n^2f(n)-(n-1)^2f(n-1)$$ $$f(n)=\frac{n-1}{n+1}\cdot f(n-1)$$

which gives a recurrance equation.

Answer 3

As suggested by @Vincent, we have $$f(n)=\frac{n-1}{n+1}f(n-1)$$ $$=\frac{(n-1)(n-2)}{(n+1)n}f(n-2)$$ $$=...=2\frac{(n-1)!}{(n+1)!}f(1)$$

$$=2\frac{1}{n(n+1)}f(1)$$ $$\implies f(2016)=\frac{2}{2017}.$$