Question

A function $f$ is defined for all positive integers and satisfies $f\left(1\right)=2005$ and $n^{2}f\left(n\right)=f\left(1\right)+f\left(2\right)+...+f\left(n\right)$ for all $n>1$. Find the value of $f\left(2004\right)$.

Answer 1

Substituting $n=2$ in the given equation we get,

$4f\left(2\right)=f\left(1\right)+f\left(2\right)$

$\Rightarrow f\left(2\right)=\frac{f\left(1\right)}{3}=\frac{f\left(1\right)}{1+2}$

Now putting $n=3$ we get,

$9f\left(3\right)=f\left(1\right)+f\left(2\right)+f\left(3\right)$

$\Rightarrow 8f\left(3\right)=\frac{4f\left(1\right)}{3}$

$\Rightarrow f\left(3\right)=\frac{f\left(1\right)}{6}=\frac{f\left(1\right)}{1+2+3}$ and so on

Thus, $f\left(2004\right)=\frac{f\left(1\right)}{1+2+3+....+2004}$ $=\frac{\left(2005\right)\times 2}{2005\times 2004}=\frac{1}{1002}$

So, the value of $f\left(2004\right)$ is $\frac{1}{1002}$.