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

Question

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)