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  • 求證:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整數(shù))

    求證:1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24(n是正整數(shù))
    數(shù)學人氣:719 ℃時間:2019-11-18 22:28:49
    優(yōu)質(zhì)解答
    證明:
    當k=1時
    1/2+1/3+1/4=13/12=26/24>25/24
    結(jié)論成立.
    假設k=n時結(jié)論成立,即
    1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24
    當k=n+1時
    由于
    9(n+1)^2=9n^2+18n+9>9n^2+18n+8=(3n+2)(3n+4)

    9(n+1)^2/[(3n+2)(3n+4)]-1>0
    左側(cè)為
    1/[(n+1)+1]+1/[(n+1)+2]+1/[(n+1)+3]+...+1/[3(n+1)+1]
    =1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{1/(3n+2)+1/(3n+3)+1/(3n+4)-1/(n+1)}
    =1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+{6(n+1)/[(3n+2)(3n+4)]-2/(3n+3)}
    =1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)+2/(3n+3)*{9(n+1)^2/[(3n+2)(3n+4)]-1}
    >1/(n+1)+1/(n+2)+1/(n+3)+...+1/(3n+1)>25/24.
    結(jié)論成立.
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