Given that $f(1)= 2013,$ find the value of $f(2013)$? Suppose that $f$ is a function  defined on the set of natural numbers such that $$f(1)+ 2^2f(2)+ 3^2f(3)+...+n^2f(n) = n^3f(n)$$ for all positive integers $n$.
 Given that $f(1)= 2013$, find the value of $f(2013)$.
 A: Introduce the auxiliary function
$$g(n):=n^2 f(n)\qquad(n\geq1)\ .$$
Then
$$n g(n)= g(1)+g(2)+\ldots+g(n)\qquad(n\geq1)$$
and therefore
$$(n+1)g(n+1)-n g(n)=g(n+1)\ ,$$
or $g(n+1)=g(n)$ for all $n\geq1$. It follows that
$$2013^2 f(2013)= g(2013)= g(1)=1^2 f(1)\ ,$$
whence $f(2013)={1\over2013}$.
A: We have $$\sum_{1\le r\le n} r^2f(r)=n^3f(n)$$
Putting $n=m, \sum_{1\le r\le m} r^2f(r)=m^3f(m)$
Putting $n=m+1, \sum_{1\le r\le m+1} r^2f(r)=(m+1)^3f(m+1)$
On subtraction, $$m^3f(m)=f(m+1)\{(m+1)^3-(m+1)^2\}$$
$$f(m+1)=f(m)\cdot\left(\frac m{m+1}\right)^2=f(m-1)\cdot\left(\frac {m(m-1)}{(m+1)m}\right)^2=\cdots = \frac{f(1)}{(m+1)^2}\text{ for }m+1\ge 1\implies m\ge 0 $$
A: Hint: Try computing some small examples first.  When $n=2$, we get that $$f(1)+2^{2}f(2)=2^{3}f(2) $$ $$\Rightarrow 2^2f(2)=f(1).$$  Using the previous case and the given equation, when $n=3,$ we have that 
$$f(1)+2^2f(2)+3^{2}f(3)=3^{3}f(3)$$ 
$$\Rightarrow f(1)+f(1)+3^{2}f(3)=3^{3}f(3),$$ 
$$\Rightarrow 18f(3)=2f(1),$$
$$\Rightarrow 3^{2}f(3)=f(1).$$   
Using the previous cases and the given equation, when $n=4$ we have 
$$f(1)+2^2f(2)+3^2f(3)+4^{2}f(4)=4^{3}f(4), $$ 
$$\Rightarrow f(1)+f(1)+f(1)+4^{2}f(4)=4^{3}f(4), $$
$$\Rightarrow 48f(4)=3f(1),$$
$$\Rightarrow 4^2f(4)=f(1).$$
Do you see a pattern?  What is happening here? Now that you have a conjecture, try to prove it.
A: This is an easy question.
Let's prove first that for every non negative integer, the following holds:
$$n^2 f(n)=f(1)$$
For $n=2$:
$$f(1)+2^{2} f(2)=2^{3} f(2)$$
$$2^2 f(2)=f(1)$$
Suppose that for every $p$ less than $n$:
$$p^2 f(p)=f(1)$$
Then by hypothesis
$$f(1)+2^2 f(2)+3^2 f(3)+..+(n-1)^2 f(n-1)+n^2 f(n)=n^3 f(n)$$
So:
$$(n^3-n^2) f(n)=(n-1) f(1)$$
Then:
$$n^2 f(n)=f(1)$$
Thus:
$$f(n)=f(1)/(n^2)$$
Applying to the $n=2013$ case:
$$f(2013)=2013/(2013^2)=1/2013$$
