1. Given the sets
$$
\begin{array}{c}
M=\{x, x y, \lg (x y)\} \\
N=\{0,|x|, y\},
\end{array}
$$

and
and $M=N$, then
$\left(x+\frac{1}{y}\right)+\left(x^{2}+\frac{1}{y^{2}}\right)+\left(x^{3}+\frac{1}{y^{3}}\right)+\cdots+\left(x^{2001}+\frac{1}{y^{2001}}\right)$ is equal to . $\qquad$
1. From the equality of sets, we know that the elements of the two sets are the same. Thus, there must be an element 0 in $M$. Also, by the definition of logarithms, $x y \neq 0$, so $x, y$ are not zero. Therefore, $\lg (x y)=0$, $x y=1$.
$$
\therefore \quad M=\{x, 1,0\}, N=\left\{0,|x|, \frac{1}{x}\right\} \text {. }
$$

From the equality of sets, we have $\left\{\begin{array}{l}x=|x|, \\ 1=\frac{1}{x} ;\end{array}\right.$ or $\left\{\begin{array}{l}x=\frac{1}{x}, \\ 1=|x|\end{array}\right.$
However, when $x=1$, it will contradict the distinctness of elements in the same set, so we only have $x=-1$, and thus $y=-1$. Therefore,
$$
\begin{array}{l}
x^{2 k+1}+\frac{1}{y^{2 k+1}}=-2 \quad(k=0,1,2, \cdots), \\
x^{2 k}+\frac{1}{y^{2 k}}=2 \quad(k=1,2, \cdots) .
\end{array}
$$

Hence, the value of the required algebraic expression is -2.