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

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But I'm lost here.

What do I do now?

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  • $\begingroup$ $f(1)+f(2)=2^2f(2)\implies f(2)=672$ and $f(1)+f(2)+f(3)=3^2f(3)$ .Continue this procedure. $\endgroup$ – Behrouz Maleki Jan 5 '17 at 12:01
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$$f(1)+\ldots+f(n)=n^2f(n)$$ $$f(1)+\ldots+f(n-1)=(n-1)^2f(n-1)$$

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.

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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.

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

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

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  • $\begingroup$ Sorry, may I ask where does the constant 2 come from? after expressing the right side into factorial form? $\endgroup$ – rosa Jun 26 '17 at 10:20
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    $\begingroup$ because $ f (2 )=1/3 f (1)= 2. 1/3.2 f (1) $ to complete denominator . $\endgroup$ – hamam_Abdallah Jun 26 '17 at 10:26
  • $\begingroup$ oh??? I overlook that one tnx :) Great Math :) $\endgroup$ – rosa Jun 26 '17 at 10:34

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