答一答：有若干个数，第一个数记为$a_{1}$，第$2$个数记为$a_{2}$，第$3$个数记为$a_{3}\ldots $第$n$个数记为$a_{n}$，若$a_{1}=-\dfrac{1}{2}$，$a_{n}=\dfrac{1}{1-a_{n-1}}(n\geqslant 2$，$n$为整数）求$a_{2}$，$a_{3}$，$a_{2006}$.

答一答：有若干个数，第一个数记为$a_{1}$，第$2$个数记为$a_{2}$，第$3$个数记为$a_{3}\ldots $第$n$个数记为$a_{n}$，若$a_{1}=-\dfrac{1}{2}$，$a_{n}=\dfrac{1}{1-a_{n-1}}(n\geqslant 2$，$n$为整数）求$a_{2}$，$a_{3}$，$a_{2006}$.

$\because a_{1}=-\dfrac{1}{2}$；

$\therefore a_{2}=\dfrac{1}{1-\left(-\dfrac{1}{2}\right)}=\dfrac{2}{3}$；

$a_{3}=\dfrac{1}{1-\dfrac{2}{3}}=3$，

$\ldots$

$-\dfrac{1}{2}$，$\dfrac{2}{3}$，$3$依次循环，

又$\because 2006\div 3=668\ldots 2$，

$\therefore a_{2006}=\dfrac{2}{3}$.