问题
解答题
已知函数f(x)=
(1)求数列{an}的通项公式an; (2)若数列{bn}满足bn=
|
答案
(1)由已知,an+1=
,所以an an+3
=1 an+1
+1,3 an
∴
+1 an+1
=3(1 2
+1 an
),1 2
∴数列{
+1 an
}是以1+1 2
=1 2
为首项,以3为公比的等比数列.3 2
∴
+1 an
=1 2
•3 n-1=3 2
,3n 2
=1 an 3n-1 2
所以an=2 3n-1
(2)bn=
anan+1•3n=1 2
=2•3n (3n-1)(3n+1-1)
-1 3n-1 1 3n+1-1
Sn=b1+b2+…+bn=
-1 31-1
+(1 32-1
-1 32-1
)+…+(1 33-1
-1 3n-1
)=1 3n+1-1
-1 2 1 3n+1-1