问题
解答题
已知数列{an}中,a1=
(1)求证:数列{bn}是等比数列; (2)求数列{an}的通项公式. |
答案
(1)证明:bn=an+1-
an=[1 2
an+(1 3
)n+1]-1 2
an=(1 2
)n+1-1 2
an,bn+1=(1 6
)n+2-1 2
an+1=(1 6
)n+2-1 2
[1 6
an+(1 3
)n+1]=1 2
•(1 2
)n+1-1 2
an-1 18
•(1 6
)n+1=1 2
•(1 3
)n+1-1 2
an=1 18
•[(1 3
)n+1-1 2
an],1 6
∴
=bn+1 bn
(n=1,2,3,…).1 3
∴{bn}是公比为
的等比数列.1 3
(2)∵b1=(
)2-1 2
a1=1 6
-1 4
•1 6
=5 6
,1 9
∴bn=
•(1 9
)n-1=(1 3
)n+1.1 3
由bn=(
)n+1-1 2
an,得(1 6
)n+1=(1 3
)n+1-1 2
an,解得an=6[(1 6
)n+1-(1 2
)n+1].1 3