14. Let $a_{1}=2006$, and for $n \geq 2$,
$$
a_{1}+a_{2}+\cdots+a_{n}=n^{2} a_{n} .
$$

What is the value of $2005 a_{2005}$ ?
14. Ans: 2
$$
a_{n}=\sum_{i=1}^{n} a_{i}-\sum_{i=1}^{n-1} a_{i}=n^{2} a_{n}-(n-1)^{2} a_{n-1} .
$$

This gives $a_{n}=\frac{n-1}{n+1} a_{n-1}$. Thus,
$$
a_{2005}=\frac{2004}{2006} a_{2004}=\cdots=\frac{(2004)(2003) \cdots(1)}{(2006)(2005) \cdots(3)} a_{1}=\frac{2}{2005}
$$