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    • 1/(1*2)+2/(1*2*3)+3/(1*2*3*4)+...+n/(1*2*3*...*(n+1))

    1/(1*2)+2/(1*2*3)+3/(1*2*3*4)+...+n/(1*2*3*...*(n+1))

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    其他回答
    第1个回答  2007-10-03
    n/(1*2*3*...*(n+1)=[1/n!-1/(n+1)!]
    所以原式＝1-1/(n+1)!
    第2个回答  2007-10-03
    an=1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
    原式＝a1+a2+a3+...+an
    =1/1-1/1*2+1/1*2-1/1*2*3+1/1*2*3-1/1*2*3*4+...+1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
    =1-1/(1*2*3*...*(n+1)).
    第3个回答  2007-10-03
    取an＝n/(1*2*3*...*(n+1))，则
    an=1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
    原式＝a1+a2+a3+...+an
    =1/1-1/1*2+1/1*2-1/1*2*3+1/1*2*3-1/1*2*3*4+...+1/(1*2*3*...*n)-1/(1*2*3*...*(n+1))
    =1-1/(1*2*3*...*(n+1)).
    第4个回答  推荐于2021-01-21
    1/(1*2)+2/(1*2*3)+3/(1*2*3*4)+...+n/(1*2*3*...*(n+1))
    =1-1/2+1/2-1/(2*3)+1/(2*3)-1/(2*3*4)+...+1/(2*3*..*n)-1/(3*4*5*....*(n+1))
    =1-1/(3*4*5*....*(n+1))本回答被提问者采纳
    第5个回答  2007-10-03
    =1/(1*2)+1/(1*2)-1/(1*2*3)+1/(1*2*3)......-1/(1*2*3*...*(n+1))=1-1/(1*2*3*...*(n+1))
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