设数列{an}中,Sn是它的前n项和,a1=4,nan+1=Sn+n(n+1)对任意n∈N*均成立.(I)求证:数列{an}是等差数列;(II)设数列{bn}满足bn+1-bn=an,其中b1=2,求

题目简介

设数列{an}中,Sn是它的前n项和,a1=4,nan+1=Sn+n(n+1)对任意n∈N*均成立.(I)求证:数列{an}是等差数列;(II)设数列{bn}满足bn+1-bn=an,其中b1=2,求

题目详情

设数列{an}中,Sn是它的前n项和,a1=4,nan+1=Sn+n(n+1)对任意n∈N*均成立.
(I)求证:数列{an}是等差数列;
(II)设数列{bn}满足bn+1-bn=an,其中b1=2,求数列{bn}的通项公式;
(III)设cn=
1
bn
,求证:c1+c2+…+cn<1.
题型:解答题难度:中档来源:丰台区二模

答案

(I)∵nan+1=Sn+n(n+1)①∴(n-1)an=Sn-1+(n-1)n(n≥2)②
①-②整理得,an+1-an=2(n≥2)
又由①,取n=1得a2-a1=2∴an+1-an=2(n∈N*)
∴数列{an}是以4为首项,2为公差的等差数列.
(II)由(I)知an=4+2(n-1)=2(n+1)
∴bn+1-bn=2(n+1)
∴(bn-bn-1)+(bn-1-bn-2)+…+(b3-b2)+(b2-b1)=2n+2(n-1)+…+2×3+2×2=n2+n-2
∴bn=n(n+1).
(III)由cn=class="stub"1
bn
得,cn=class="stub"1
n(n+1)
=class="stub"1
n
-class="stub"1
n+1

∴c1+c2+…+cn=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

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