问题
解答题
设数列{an}(n∈N*)的前n项的和为Sn,满足a1=1,
(1)求证:Sn=(2-
(2)求数列{an}的通项公式. |
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
(n∈N*).1 2n
所以
-S2 a2
=S1 a1
,1 2
-S3 a3
=S2 a2
;1 4
-S4 a4
=S3 a3
;1 8
…
-Sn an
=Sn-1 an-1
;1 2n-1
将n-1个式子相加可得:
-Sn an
=S1 a1
+1 2
+ 1 22
+…+1 23
,1 2n-1
所以
=1+Sn an
+1 2
+1 22
+…+1 23
=1 2n-1
=2-1- 1 2n-1 1- 1 2
;1 2n-1
∴Sn=(2-
)an;1 2n-1
(2)因为Sn=(2-
)an;1 2n-1
所以Sn-1=(2-
)an-1;(n≥2)1 2n-2
所以an=(2-
)an-(2-1 2n-1
)an-1;可得1 2n-2
an =an-1,1 2
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.