已知数列{an}中,a1=1,且点P(an,an+1)(n∈N*)在直线x-y+1=0上.(1)求数列{an}的通项公式;(2)若函数f(n)=1n+a1+1n+a2+1n+a3+…+1n+an(n∈

题目简介

已知数列{an}中,a1=1,且点P(an,an+1)(n∈N*)在直线x-y+1=0上.(1)求数列{an}的通项公式;(2)若函数f(n)=1n+a1+1n+a2+1n+a3+…+1n+an(n∈

题目详情

已知数列{an}中,a1=1,且点P(an,an+1)(n∈N*)在直线x-y+1=0上.
(1)求数列{an}的通项公式;
(2)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.
题型:解答题难度:中档来源:不详

答案

(1)由点P(an,an+1)在直线x-y+1=0上,
即an+1-an=1,且a1=1,数列{an}是以1为首项,1为公差的等差数列,
an=1+(n+1)•1=n(n≥2),a1=1同样满足,
所以an=n.
(2)f(n)=class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
2n
f(n+1)=class="stub"1
n+2
+class="stub"1
n+3
+class="stub"1
n+4
+class="stub"1
2n+1
+class="stub"1
2n+2
f(n+1)-f(n)=class="stub"1
2n+1
+class="stub"1
2n+2
-class="stub"1
n+1
>class="stub"1
2n+2
+class="stub"1
2n+2
-class="stub"1
n+1
=0

所以f(n)是单调递增,
故f(n)的最小值是f(2)=class="stub"7
12

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