已知数列{an}中,a1=1,且点p(an,a(n+1))(n属于N)在直线x-y+1=0上,若函数f(n)=1/(n+
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/10/05 22:24:03
已知数列{an}中,a1=1,且点p(an,a(n+1))(n属于N)在直线x-y+1=0上,若函数f(n)=1/(n+a1)+1/(n+a2)+1/(n+a3)+…+1/(n+an)(n∈N,且n≥2),求函数f(n)的最小值;
点p(an,a(n+1))(n属于N)在直线x-y+1=0上,
an+1-an=1
数列{an}中,a1=1.d=1
所以an=a1+(n-1)*1=n
∴f(n)=1/(n+a1)+1/(n+a2)+1/(n+a3)+…+1/(n+an)(n∈N,且n≥2)
==>f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+n)
=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(2n)
∴f(n+1)=1/(n+1+1)+1/(n+1+2)+1/(n+1+3)+...+1/(n+1+n+1)
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)
∴当n≥2时,有
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)>1/(2n+2)+1/(2n+2)-1/(n+1)=0
即f(n+1)>f(n)
则当n≥2时,f(n)>f(n-1)>f(n-2)>...>f(2)
∴n=2时,
f(n)取最小值f(2)=1/(2+1)+1/(2+2)=7/12
an+1-an=1
数列{an}中,a1=1.d=1
所以an=a1+(n-1)*1=n
∴f(n)=1/(n+a1)+1/(n+a2)+1/(n+a3)+…+1/(n+an)(n∈N,且n≥2)
==>f(n)=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(n+n)
=1/(n+1)+1/(n+2)+1/(n+3)+…+1/(2n)
∴f(n+1)=1/(n+1+1)+1/(n+1+2)+1/(n+1+3)+...+1/(n+1+n+1)
=1/(n+2)+1/(n+3)+...+1/(2n)+1/(2n+1)+1/(2n+2)
∴当n≥2时,有
f(n+1)-f(n)=1/(2n+1)+1/(2n+2)-1/(n+1)>1/(2n+2)+1/(2n+2)-1/(n+1)=0
即f(n+1)>f(n)
则当n≥2时,f(n)>f(n-1)>f(n-2)>...>f(2)
∴n=2时,
f(n)取最小值f(2)=1/(2+1)+1/(2+2)=7/12
已知数列{an}中,a1=1,且点p(an,an+1)(n属于正整数)在直线x-y+1=0上
已知数列an中,a1=1,前n项和为Sn,且点P(an,an+1)(n属于N*)在直线x-y+1=0上,则1/S1+1/
已知数列{an}中,a1=1,且点p(an,a(n+1))(n∈N*)在一次函数y=x+1上 (1)求数列{an}的通项
已知数列{an}中,a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则1S
已知在数列|an|中,a1=1,且点(an,an+1)(n∈N*)在函数f(x)=x+2的图像上
已知数列{an}中,a1=-2008点P(an,a(n+1))在直线x-y+3=0上,
已知数列{an},a1=1,前n项和为Sn,且点P(an,an+1)(n∈N*)在直线x-y+1=0上,则1S1+1S2
已知数列{an}中,a1=2,且点p(an,an+1)(n∈N*)在斜率为1,纵截距为2的直线上
已知在数列﹛an﹜中,a1=1,且点(an,a(n+1))(n∈N+)在函数f(x)=x+2的图像上.
在数列{An}中,a1=2,且点P(an,an-1)在直线2X-Y=0上,1求数列{An}通项公式 2设bn=n/an,
已知数列an中a1=1/2点(n,2an+1-an)在直线y=x上其n=1,2,3……(n,2an+1-an)中的an+
已知数列{an}中a1=6,且an-an-1=(an-1/n)+n+1(n属于N*,n≥2),求an