若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f
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若f(n)=[1/(n+1)]+[1/(n+2)]+[1/(n+3)]+``` ```+(1/2n),则f(n+1)-f(n)=
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[1/(2n+1)]-[1/(2n+2)]
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