6. $a_{n} \in \mathbf{R}^{+}$ and $a_{1}=1, a_{2}=10, a_{n}^{2} a_{n-2}=10 a_{n-1}^{3}$, $n=3,4, \cdots$. Find $a_{n}$. 1987, China Mathematical Olympiad Training Team Selection Test)
(Solution: Taking logarithms on both sides, we get $\lg a_{n}=\frac{3}{2} \lg a_{n-1}$ $-\frac{1}{2} \lg a_{n-2}+\frac{1}{2}$
(1). From the characteristic equation: $x^{2}=\frac{3}{2} x-$ $\frac{1}{2}$, we get $x_{1}=1, x_{2}=\frac{1}{2}$. The particular solution is $d_{n}=n d$. Substituting into (1), we get $(n+2) d=\frac{3}{2}(n+1) d-\frac{1}{2} d_{n}+\frac{1}{2} \Rightarrow d=1$. Therefore, 
$d_{n}=n$. Thus, the general solution is $\lg a_{n}=c_{1}+c_{2}\left(\frac{1}{2}\right)^{n-1}+n$. Using the initial conditions, we get $c_{1}=-1, c_{2}=0$. Hence, $\lg a_{n}=n-1, a_{n}$ $=10^{n-1}$. $)$