Sequences and Sums There is a list of numbers $a_{1} , a_{2} , …, a_{2010}$ . For $1 \leq n \leq 2010$, where $n$ is positive
integer, let $a_1+a_2+ \ldots +a_n = S_n$ . If $a_1 = 2010$ and $S_n = a_nn^2$ for all n, what is
the value of $a_{2010}$ ?
I've been trying to manipulate the formula but I cant seem to find a good relationship between $a_1$  and $a_{2010}$ like
$$
a_{2010} = \frac{a_1 +a_2 ... +a_{2010}}{2010^2}
$$ 
Then tried to use the definition $S_n = a_nn^2 $ over and over again but I cant find a good formula.

We have $a_n\cdot n^2=a_1+a_2+\ldots+a_n=S_n$. Then
$$
\begin{array}{cc}
\begin{array}{rlrl}
\frac{\quad}{\quad}a_1\cdot 1^2=&2010\\
\frac{\quad}{\quad}a_2\cdot 2^2=&a_1+a_2\\
\frac{\quad}{\quad}a_3\cdot 3^2=&a_1+a_2+a_3\\
\frac{\quad}{\quad}a_4\cdot 4^2=&a_1+a_2+a_3+a_4\\
\end{array}
&
\begin{array}{rl}
\implies a_1=&2010\\
\implies a_2 =& \frac{1}{2^2-1}2010\\
\implies a_3=&\frac{1}{3^2-1}2010+\frac{1}{3^2-1}\frac{1}{2^2-1}2010\\
\implies a_4=&\frac{1}{4^2-1}2010+\frac{1}{4^2-1}\frac{1}{3^2-1}2010+\frac{1}{4^2-1}\frac{1}{3^2-1}\frac{1}{2^2-1}2010\\
\end{array}
\end{array}
$$
Now is easy to see 
$$
a_n=2010\left(\frac{1}{n^2-1}+\frac{1}{n^2-1}\cdot\frac{1}{(n-1)^2-1}+\ldots+\frac{1}{n^2-1}\cdot\frac{1}{(n-1)^2-1}\cdot\ldots\cdot\frac{1}{2^2-1}\right)
$$
