UK 1996/2

[tanmoy]

A function $f(x)$ defined on the positive integers satisfies $f(1) = 1996$ and $f(1)+f(2)+\ldots + f(n) = n^2f(n)\qquad (n > 1)$. Calculate $f(1996)$.

[tanmoy]

Hints:

Induction kills it

[Nirjhor]

Let $f(1)=C$ and $P(n)~\Rightarrow ~f(1)+\cdots+f(n)=n^2 f(n)$. Then rearranging $P(n)-P(n-1)$ gives $\dfrac{f(n)}{f(n-1)}=\dfrac{n-1}{n+1}$. Multiplying similar expressions leads to \[\prod_{k=2}^n \dfrac{f(k)}{f(k-1)}=\prod_{k=2}^n \dfrac{k-1}{k+1}~\Rightarrow ~\dfrac{f(n)}{f(1)}=\dfrac{2}{n(n+1)}~\Rightarrow ~f(n)=\dfrac{2C}{n(n+1)}.\] In this case, $f(1996)=\dfrac{2}{1997}$.