观察下列等式①1⼀√2+1=√2-1⼀(√2+1)(√2-1)=√2-1, ②1⼀√3+√2=√3-√2⼀(√3+√2)(√3-√2

2025-01-02 12:51:34
推荐回答(1个)
回答1:

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