f(n)=1/n+1+1/n+2+/1n+3+.+1/2n(n包涵正整数那么f(n+1)-f(n)=
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f(n)=1/n+1+1/n+2+/1n+3+.+1/2n(n包涵正整数那么f(n+1)-f(n)=
f(n)=1/(n+1)+1/(n+2)+1/(n+3)+.+1/2n
f(n+1)=1/(n+2)+1/(n+3)+.+1/(2n+1)+1/(2n+2)
f(n+1)-f(n)=[1/(n+2)+1/(n+3)+.+1/(2n+1)]-[1/(n+1)+1/(n+2)+1/(n+3)+.+1/2n]
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/(2n+1)(2n+2)
f(n+1)=1/(n+2)+1/(n+3)+.+1/(2n+1)+1/(2n+2)
f(n+1)-f(n)=[1/(n+2)+1/(n+3)+.+1/(2n+1)]-[1/(n+1)+1/(n+2)+1/(n+3)+.+1/2n]
=1/(2n+1)+1/(2n+2)-1/(n+1)
=1/(2n+1)-1/(2n+2)
=1/(2n+1)(2n+2)
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