Solving for $f(2004)$ in a given functional equation

Question

Given that $$f(1)=2005$$ and $$f(1)+f(2)+...f(n) = n^{2}f(n)$$ for all $n>1$. Determine the value of $f(2004)$.

My progress:

I first substituted $n-1$ into the equation to get $f(1)+f(2)+...+f(n-1) = (n-1)^{2}f(n-1)$ and then I subtracted it from the first equation to get $$f(n) = n^{2}f(n)-(n-1)^{2}f(n-1)$$. I got a little stuck so I checked the answers for a hint and it said that: $$f(n) = \frac{(n-1)^{2}}{n^{2}-1}f(n-1)$$. How is this so? I don't understand the last part.

Answer 1

You have $$\begin{align}f(1)+f(2)+...+f(n-1)+f(n) &= n^{2}f(n)\end{align}$$

send that $f(n)$ to right hand side and get $$\begin{align}f(1)+f(2)+...+f(n-1) &= n^{2}f(n)-f(n)\\~\\&=(n^2-1)f(n)\end{align}$$

Next observe that left hand side becomes $(n-1)^2f(n-1)$

Answer 2

Because $n^2-1 = (n-1)(n+1)$, you can get \begin{align} f(n) = \frac{n-1}{n+1}f(n-1) \end{align} The remaining is simple.

Answer 3

Then... from this equation you can easily get that:

$f(n) = (2 / (n(n+1))).f(1)$

How? Just express f(n) by f(n-1), then f(n-1) by f(n-2) and so on until you reach the bottom.

Answer 4

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

and similarly

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

Subtracting the second from the first we get $$ f(n)=n^2f(n)-(n-1)^2f(n-1) $$i.e $$ f(n)=n^2f(n)-(n-1)^2f(n-1) $$ or $$ 0=(n^2-1)f(n)-(n-1)^2f(n-1) $$ i.e $$ f(n)=\frac{(n-1)^2f(n-1))}{(n^2-1)} $$ $$ f(n)=\frac{n-1}{n+1}f(n-1) $$ Replacing $n$ by $n-1$ and so on we get the equation $$ f(n)=\frac{n-1}{n+1}\frac{n-2}{n}\frac{n-3}{n-1}\dots\frac{3}{5}\frac{2}{4}\frac{1}{3}f(1) $$ Cancelling out the common terms we get $$ f(n)=\frac{2*f(1)}{(n+1)(n)} $$ $$ \therefore f(2004)=\frac{2*2005}{2005*2004}=\frac{1}{1002} $$