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
人气：469 ℃ 时间：2020-01-30 23:20:45
解答
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)...
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