已知数列{an}中,a1=1,且点P(an,an+1 【注：n+1为a的下标】)（n属于正整数）在直线X-Y+2=0上.若函数f(n)=2/(2n+a1)+2/(2n+a2)+2/(2n+a3)+…+2/（2n+an）（n∈N）,求函数f（n）的最小值；
问题描述：
已知数列{an}中,a1=1,且点P(an,an+1 【注：n+1为a的下标】)（n属于正整数）在直线X-Y+2=0上.若函数f(n)=2/(2n+a1)+2/(2n+a2)+2/(2n+a3)+…+2/（2n+an）（n∈N）,求函数f（n）的最小值；
答
2/3
答
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