1. Simplify the expression $A=1-a+a^{2}-a^{3}+\ldots+a^{1996}-a^{1997}+\frac{a^{1998}}{1+a}$ and calculate its value for $a=-\frac{1996}{1997}$.
Solution:

$$
\begin{aligned}
A & =\frac{1+a-a(1+a)+a^{2}(1+a)-a^{3}(1+a)+\ldots+a^{1996}(1+a)-a^{1997}(1+a)+a^{1998}}{1+a}= \\
& =\frac{1+a-a-a^{2}+a^{2}+a^{3}-a^{3}+\ldots+a^{1997}-a^{1997}-a^{1998}+a^{1998}}{1+a}=\frac{1}{1+a}
\end{aligned}
$$

$$
\text { 3a } a=-\frac{1996}{1997}, A=\frac{1}{1-\frac{1996}{1997}}=1997
$$