观察下列各运算：（√2-1）（√2+1）=1

[1/（1+√2）+1/（√2+√3）+1/（√3+√4）+……+1/（√2001+√2002）+1/（√2002+√2003）]
=(√2-1)/[(√2+1)(√2-1)]+(√3-√2)/[(√3+√2)(√3-√2)]+(√4-√3)/[(√4+√3)(√4-√3)]+...+(√2003-√2002)/[(√2002+√2003)(√2003-√2002)]
=√2-1+√3-√2+√4-√3+...+√2003-√2002
=√2003-1

将题中的每一项都分母有理化。然后每隔两项可相互抵消。
最后得：根号2003减1