已知数列{an}满足递推式:an+1-2an=an-2an-1(n≥2,n∈N),a1=1,a2=3.(Ⅰ)若bn=11+an,求bn+1与bn

2025-06-22 06:44:44
推荐回答(1个)
回答1:

(Ⅰ)解:an+1-

2
an
=an-
2
an-1
=…=a2-
2
a1
=3-2=1?an+1-
2
an
=1

bn=
1
1+an
?an=
1
bn
-1
代入①式得
1
bn+1
-1-
2
1
bn
-1
=1?
1-bn+1
bn+1
-
2bn
1-bn
=1

bn+1=-
1
2
bn+
1
2

(Ⅱ)证明:
1
1+an
=
1
3
[1-(-
1
2
)
n
]?an+1=
3
1-(-
1
2
)
n
?|an-2|=|
3
1-(-
1
2
)
n
-3|=
3
|(-2)n-1|

对n分奇数与偶数讨论:|a2k-1-2|=
3
22k-1+1
,|a2k-2|=
3
22k-1

|a