如果f(n)=1/(n+1)+1/(n+2)+```1/2n (n属于N*) 那么f(n+1)-f(n)=

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如果f(n)=1/(n+1)+1/(n+2)+```1/2n (n属于N*) 那么f(n+1)-f(n)=
如果f(n)=1/(n+1)+1/(n+2)+```1/2n (n属于N*) 那么f(n+1)-f(n)=

如果f(n)=1/(n+1)+1/(n+2)+```1/2n (n属于N*) 那么f(n+1)-f(n)=
f(n)有n项,则
f(n+1)-f(n)
=[(1/(n+2))+(1/(n+3))+···+(1/(2n))+(1/(2n+1))+(1/(2n+2))]
-[(1/(n+1))+(1/(n+2))+···+(1/(2n))]
=[(1/(2n+1))+(1/(2n+2))]-(1/(n+1))
=(1/(2n+1))-(1/(2n+2))
=1/(2(n+1)(2n+1))
祝你学习天天向上,加油~

-1/(n+1)+1/(2n+1)+1/(2n+2)

f(n)={n+1}--->{2n}.====>f(n+1)={n+2}---->{2n},{2n+1},{2n+2}.===>f(n+1)-f(n)={2n+2}+{2n+1}-{n+1}=1/[(2n+1)(2n+2)].