Question

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

Answer 1

$n^{2}f\left(n\right)=f\left(1\right)+f92)+f(3)+...+f(n)$

$(n+1{)}^{2}f(n+1)=f(1)+f(2)+f(3)+...+f(n+1)$

$⟹(n+1{)}^{2}f(n+1)-n^{2}f(n)=f(n+1)$

$⟹f(n+1)=\frac{n^2}{n^2+2n}f\left(n\right)$

$⟹f(n+1)=\frac{n}{n+2}f\left(n\right)$

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

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

and so on...

Finally..

$⟹f(n+1)=\frac{n}{n+2}\frac{n-1}{n+1}\frac{n-2}{n}\frac{n-3}{n-1}...\frac{3}{5}\frac{2}{4}\frac{1}{3}f\left(1\right)$

Cancelling the terms and proceeding we get-

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

For $n=2003$

$f\left(2004\right)=\frac{2\times 2005}{2005\times 2004}$

$⟹f\left(2004\right)=\frac{1}{1002}$