TITLE: Given that $f(1)= 2013,$ find the value of $f(2013)$?
QUESTION [7 upvotes]: 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 $f(2013)$.

REPLY [2 votes]: 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, $$m^3f(m)=f(m+1)\{(m+1)^3-(m+1)^2\}$$
$$f(m+1)=f(m)\cdot\left(\frac m{m+1}\right)^2=f(m-1)\cdot\left(\frac {m(m-1)}{(m+1)m}\right)^2=\cdots = \frac{f(1)}{(m+1)^2}\text{ for }m+1\ge 1\implies m\ge 0 $$