已知数列{an}满足:a1=2,an+1=an+1n(n+1),n∈N*.(I)求数列{an}的通项公式an;(II)设bn=n2nan(n∈N*),求数列{bn}的前n项和Sn.-数学

题目简介

已知数列{an}满足:a1=2,an+1=an+1n(n+1),n∈N*.(I)求数列{an}的通项公式an;(II)设bn=n2nan(n∈N*),求数列{bn}的前n项和Sn.-数学

题目详情

已知数列{an}满足:a1=2,an+1=an+
1
n(n+1)
,n∈N*
(I)求数列{an}的通项公式an
(II)设bn=
n
2n
an
(n∈N*),求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:通州区一模

答案

(I)∵an+1=an+class="stub"1
n(n+1)

an+1-an=class="stub"1
n
-class="stub"1
n+1

a2-a1=1-class="stub"1
2
a3-a2=class="stub"1
2
-class="stub"1
3
,…,an-an-1=class="stub"1
n-1
-class="stub"1
n

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

∵a1=2,∴an=3-class="stub"1
n

(II)bn=class="stub"n
2n
an
=(3n-1)•class="stub"1
2n

∴Sn=2•class="stub"1
2
+5•class="stub"1
22
+…+(3n-1)•class="stub"1
2n
①,
class="stub"1
2
Sn=2•class="stub"1
22
+5•class="stub"1
23
+…+(3n-4)•class="stub"1
2n
+(3n-1)•class="stub"1
2n+1
②,
①-②可得class="stub"1
2
Sn=2•class="stub"1
2
+3•class="stub"1
22
+…+3•class="stub"1
2n
-(3n-1)•class="stub"1
2n+1

class="stub"1
2
Sn=class="stub"5
2
-3•class="stub"1
2n
-(3n-1)•class="stub"1
2n+1

∴Sn=5-class="stub"3
2n-1
-class="stub"3n-1
2n

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