13问答网 > 1⼀(2倍根号1+根号2)+1⼀（3倍根号2+2倍根号3）+1⼀（4倍根号3+3倍根号4）+···+1⼀(100倍根号99+99倍根号10

1⼀(2倍根号1+根号2)+1⼀（3倍根号2+2倍根号3）+1⼀（4倍根号3+3倍根号4）+···+1⼀(100倍根号99+99倍根号10

2025-03-02 08:53:03

推荐回答（1个）

回答1：

考察一般项：
an=1/[(n+1)√n+n√(n+1)]
=[(n+1)√n-n√(n+1)]/[n(n+1)^2-(n+1)n^2]
=[(n+1)√n-n√(n+1)]/(n^3+2n^2+n-n^3-n^2)
=[(n+1)√n-n√(n+1)]/[n(n+1)]
=1/√n-1/√(n+1)

1/(2√1+√2)+1/(3√2+2√3)+...+1/(100√99+99√100)
=1-1/√2+1/√2-1/√3+1/√3-1/√4+...+1/√98-1/√99+1/√99-1/√100
=1-1/√100
=1-1/10
=9/10