59. Determine the value of
$$
1 \cdot 2-2 \cdot 3+3 \cdot 4-4 \cdot 5+\cdots+2001 \cdot 2002
$$
Solution: 2004002 . Rewrite the expression as
$$
\begin{array}{l} 
2+3 \cdot(4-2)+5 \cdot(6-4)+\cdots+2001 \cdot(2002-2000) \\
= 2+6+10+\cdots+4002 .
\end{array}
$$

This is an arithmetic progression with $(4002-2) / 4+1=1001$ terms and average 2002 , so its sum is $1001 \cdot 2002=2004002$.