设an=1+1/2+1/3+.+1/(3n-1)则a(n+1)-an=

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设an=1+1/2+1/3+.+1/(3n-1)则a(n+1)-an=
设an=1+1/2+1/3+.+1/(3n-1)则a(n+1)-an=

设an=1+1/2+1/3+.+1/(3n-1)则a(n+1)-an=
a(n+1) = 1+1/2+1/3+.+1/(3n-1)+...+1/[3(n+1)-1]
= 1+1/2+1/3+.+1/(3n-1)+...+1/(3n+2)
= 1+1/2+1/3+.+1/(3n-1)+1/3n+1/(3n+1)+1/(3n+2)
an = 1+1/2+1/3+.+1/(3n-1)
a(n+1) - an = 1/3n + 1/(3n+1) + 1/(3n+2)

一楼解错了。应该是这样的:
an=1+1/2+1/3+..+1/(3n-1)
a(n+1)=1+1/2+1/3+...+1/[3(n+1)-1]=1+1/2+1/3+...+1/(3n+2)
a(n+1)-an=1/(3n)+1/(3n+1)+1/(3n+2)

an =1+1/2+1/3+....+1/(3n-1)
a(n+1)=1+1/2+1/3+....+1/(3n-1)+1/(3n)
so a(n+1)-an=1/3n

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