x=an
y=a(n+1)代入直线方程：
an-a(n+1)+2=0
a(n+1)-an=2,为定值.
a1=1
数列{an}是以1为首项,2为公差的等差数列.
an=1+2(n-1)=2n-1
f(n+1)-f(n)=[2/(2n+2+a1)+2/(2n+2+a2)+...+2/(2n+2+an)+2/(2n+2+a(n+1)]-[2/(2n+a1)+2/(2n+a2)+...+2/(2n+an)]
=[2/(2n+3)+2/(2n+5)+...+2/(2n+(2n-1))+2/(2n+2n-1+2)+2/(2n+2+2n+2-1)]-[2/(2n+1)+2/(2n+3)+...+2/(2n+2n-1)]
=2/(4n+1)+2/(4n+3)-2/(2n+1)
=2[(4n+3)(2n+1)+(4n+1)(2n+1)-(4n+1)(4n+3)]/[(4n+1)(4n+3)(2n+1)]
=2/[(4n+1)(4n+3)(2n+1)]>0
f(n+1)>f(n)
f(n)单调递增,当n=1时,f(n)有最小值f(n)min=f(1)=2/(2×1+2×1-1)=2/3
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回答时间：2025-02-03 14:40
