设数列{an}(n∈N*)的前n项的和为Sn,满足a1=1,Sn+1an+1-Snan=12n(n∈N*).(1)求证:Sn=(2-12n-1)an;(2)求数列{an}的通项公式.-数学

题目简介

设数列{an}(n∈N*)的前n项的和为Sn,满足a1=1,Sn+1an+1-Snan=12n(n∈N*).(1)求证:Sn=(2-12n-1)an;(2)求数列{an}的通项公式.-数学

题目详情

设数列{an}(n∈N*)的前n项的和为Sn,满足a1=1,
Sn+1
an+1
-
Sn
an
=
1
2n
(n∈N*).
(1)求证:Sn=(2-
1
2n-1
)an
(2)求数列{an}的通项公式.
题型:解答题难度:中档来源:不详

答案

(1)证明:数列{an}(n∈N*)的前n项的和为Sn,满足a1=1,
Sn+1
an+1
-
Sn
an
=class="stub"1
2n
(n∈N*).
所以
S2
a2
-
S1
a1
 =class="stub"1
2

 
S3
a3
-
S2
a2
=class="stub"1
4

S4
a4
-
S3
a3
=class="stub"1
8


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

将n-1个式子相加可得:
Sn
an
-
S1
a1
=class="stub"1
2
+class="stub"1
22
+ class="stub"1
23
+…+class="stub"1
2n-1

所以
Sn
an
=1+class="stub"1
2
+class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n-1
=
1-class="stub"1
2n-1
1-class="stub"1
2
=2-class="stub"1
2n-1

∴Sn=(2-class="stub"1
2n-1
)an;
(2)因为Sn=(2-class="stub"1
2n-1
)an;
所以Sn-1=(2-class="stub"1
2n-2
)an-1;(n≥2)
所以an=(2-class="stub"1
2n-1
)an-(2-class="stub"1
2n-2
)an-1;可得class="stub"1
2
an =an-1

因为a2=2,当n=1时,满足数列{an}是等比数列公比为2.
所以an=2n-1.

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