已知正项数列{an}满足:an-an-1=1,(n∈N+,n≥2),且a1=4.(1)求{an}的通项公式;(2)求证1a1+1a2+…+1an<1(n∈N+)-数学

题目简介

已知正项数列{an}满足:an-an-1=1,(n∈N+,n≥2),且a1=4.(1)求{an}的通项公式;(2)求证1a1+1a2+…+1an<1(n∈N+)-数学

题目详情

已知正项数列{an}满足:
an
-
an-1
=1,(n∈N+,n≥2),且a1=4.
(1)求{an}的通项公式;
(2)求证
1
a1
+
1
a2
+…+
1
an
<1(n∈N+
题型:解答题难度:中档来源:不详

答案

(1)由题意可知数列{
an
}
是等差数列,首项是2,公差为1;
an
=2+(n-1)×1=n+1

∴an=(n+1)2
(2)证明:class="stub"1
ak
=class="stub"1
(k+1)2
<class="stub"1
k(k+1)
=class="stub"1
k
-class="stub"1
k+1

class="stub"1
a1
+class="stub"1
a2
+…+class="stub"1
an
<1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
=1-class="stub"1
n+1
<1

更多内容推荐