求数列an=(n+1)⼀n的前n项和

2025-05-11 09:31:25
推荐回答(2个)
回答1:

an=1/n(n+1)=[1/n-1/(n+1)]
Sn=a1+a2+...+an
=[1-1/2+1/2-1/3+...+1/n-1/(n+1)]
=[1-1/(n+1)]
=n/(n+1)

回答2:

Sn=n+〔n!+(n-1)!+…+3!+2!+1!〕/n!