解:(1)规律为:1/[√n+√(n-1)]=√n-√(n-1)(2)1/(√2+1)+1/(√3+√2+1/(√4+√3)+……+1/(√2001+√2002)=√2-1+√3-√2+√4-√3+……+√2002-√2001=√2002-1 [1/(√2+1)+1/(√3+√2+1/(√4+√3)+……+1/(√2001+√2002)](√2002+1)=(√2002-1)(√2002+1)=2002-1=2001