
求解,明天考试了
1个回答
2014-03-28
展开全部
an=1/[(3n-2)(3n+1)]=(1/3)[1/(3n-2)-1/(3n+1)]
Sn=(1/3)[1-/4+1/4-1/7+1/7-1/10+……+1/(3n-3)-1/(3n-2)+1/(3n-2)-1/(3n+1)]
=(1/3)[1-1/(3n+1)]
=n/(3n+1) 求采纳
Sn=(1/3)[1-/4+1/4-1/7+1/7-1/10+……+1/(3n-3)-1/(3n-2)+1/(3n-2)-1/(3n+1)]
=(1/3)[1-1/(3n+1)]
=n/(3n+1) 求采纳
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