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数列求和习题1/(1+2)+1/(1+2+3)+~+1/(1+2+3+~+n+1)
人气:391 ℃ 时间:2020-06-23 10:16:44
解答
1/(1+2+3+...+n)=2/[(n+1)*n]=2*(1/n-1/(n+1));
所以1/(1+2)+1/(1+2+3)+……+1/【1+2+3+…….+(n+1)】=2(1/2-1/3+1/3-1/4+...+1/(n+1)-1/(n+2))=2*(1/2-1/(n+2))=1-2/(n+2)=n/(n+2)
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