# let $S_n=a_1+a_2+…+a_n$ If $a_1=2010$ and $S_n=n^2a_n$ for all $n$, what is the value of $a_{2010}$?

There is a list of numbers $$a_1 , a_2,a_3,…,a_{2010}$$. For $$1\le n \le 2010$$, where $$n$$ is a positive integer, let $$S_n=a_1+a_2+…+a_n$$ If $$a_1=2010$$ and $$S_n=n^2a_n$$ for all $$n$$, what is the value of $$a_{2010}$$?

Im not sure how to solve this, I think you could derive a formula of some sort that could be used to solve it but im not sure how to do that.

Hints aswell as solutions would be appreciated

taken from the 2010 IWYMIC https://chiuchang.org/wp-content/uploads/sites/2/2018/02/2010-IWYMIC-Individual.x17381.pdf

Hint. Notice that $$a_{n+1} = S_{n+1} - S_n$$, therefore $$a_{n+1} = (n+1)^2 a_{n+1} - n^2 a_n$$ from which we get that $$a_{n+1} = \frac n {n+2}\, a_n$$ From this, can you find $$a_n$$ in terms of $$a_1$$?
$$S_1=2010$$ and $$S_n = n^2(S_n-S_{n-1})$$ imply $$S_n = \frac{n^2}{n^2-1} S_{n-1} = S_1\prod_{k=2}^{n}\frac{k^2}{k^2-1}$$ so $$a_{2010} = \frac{S_{2010}}{2010^2} = \frac{1}{2010}\prod_{k=2}^{2010}\frac{k\cdot k}{(k-1)\cdot(k+1)}=\frac{2010!\cdot2010!}{2010\cdot2009!\cdot2011!/2}=\frac{2}{2011}.$$