设数列{an}的前n项的和Sn与an的关系是Sn=-an+1-12n,n∈N*.(1)求证:数列{2nan}为等差数列,并求数列{an}的通项;(2)求数列{Sn}的前n项和Tn.-数学

题目简介

设数列{an}的前n项的和Sn与an的关系是Sn=-an+1-12n,n∈N*.(1)求证:数列{2nan}为等差数列,并求数列{an}的通项;(2)求数列{Sn}的前n项和Tn.-数学

题目详情

设数列{an}的前n项的和Sn与an的关系是Sn=-an+1-
1
2n
,n∈N*
(1)求证:数列{2nan}为等差数列,并求数列{an}的通项;
(2)求数列{Sn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,s1=-a1+1-class="stub"1
2
a1=class="stub"1
4
…(1分),
n≥2时,由Sn-Sn-1=-an+an-1+class="stub"1
2n

2nan-2n-1an-1=class="stub"1
2

∴数列{2nan}为等差数列,…(3分)
2nan=2×a1+(n-1)×class="stub"1
2
an=class="stub"n
2n+1
.…(6分)
(2)由(1)得Sn=1-class="stub"n+2
2n+1

∴Tn=n-(class="stub"3
22
+class="stub"4
23
+…+class="stub"n+2
2n+1
),①
class="stub"1
2
Tn
=class="stub"1
2
n
-(class="stub"3
23
+class="stub"4
24
+…+class="stub"n+2
2n+2
),②
①-②得class="stub"1
2
Tn
=class="stub"1
2
n
-(class="stub"3
4
+class="stub"1
23
+class="stub"1
24
+…+class="stub"1
2n+1
-class="stub"n+2
2n+2

=class="stub"1
2
n
-class="stub"3
4
-
class="stub"1
8
(1-class="stub"1
2n-1
)
1-class="stub"1
2
+class="stub"n+2
2n+2

=class="stub"1
2
n
-1+class="stub"1
2n+1
+class="stub"2n+4
2n+1
.…(9分)
∴Tn=n-2+class="stub"2n+5
2n
.…(12分)

更多内容推荐