Question

高中数学，An=3n-2，求数列{1/AnAn+1}的前n项和

n和n+1是下脚标，应该都明白吧

Answer 1

考察一般项第k项：
1/[aka(k+1)]=1/[(3k-2)(3(k+1)-2)]=1/[(3k-2)(3k+1)]=(1/3)[1/(3k-2)-1/(3k+1)]

1/(a1a2)+1/(a2a3)+...+1/[ana(n+1)]
=(1/3)[1/1-1/4+1/4-1/7+...+1/(3n-2)-1/(3n+1)]
=(1/3)[1-1/(3n+1)]
=n/(3n+1)

Answer 2

Sn=(1-1/4+1/4-1/7+1/7-1/10+......+1/(3n-2)-1/(3n+1))/3
=(1-1/(3n+1))/3=n/(3n+1)