Problem (Algebra):
Consider the sequence $t_{1}=1, t_{2}=-1$ and $t_{n}=\left(\frac{n-3}{n-1}\right) t_{n-2}$ where $n \geq 3$. What is the value of $t_{1998}$ ?

Solution:
Calculating some terms, $t_{1}=1, t_{2}=-1, t_{3}=0, t_{4}=\frac{-1}{3}, t_{5}=0, t_{6}=\frac{-1}{5}$ etc.

By pattern recognition, $t_{1998}=\frac{-1}{1997}$.

$$
\begin{aligned}
t_{1998} & =\frac{1995}{1997} t_{1996}=\frac{1995}{1997} \times \frac{1993}{1995} t_{1994} \\
& =\frac{1995}{1997} \cdot \frac{1993}{1995} \cdot \frac{1991}{1993} \cdots \frac{3}{5} \cdot \frac{1}{3} t_{2} \\
& =\frac{-1}{1997}
\end{aligned}
$$

Final answer: $\frac{-1}{1997}$