题目内容
已知f(x)=x2+2x,数列{an}满足a1=3,an+1=f′(an)-n-1,数列{bn}满足b1=2,bn+1=f(bn).(1)求证:数列{an-n}为等比数列;
(2)令cn=
1 |
an-n-1 |
2 |
3 |
(3)求证:
1 |
3 |
1 |
1+b1 |
1 |
1+b2 |
1 |
1+bn |
1 |
2 |
分析:(1)先求出函数f(x)的导函数,代入an+1=f′(an)-n-1可得an+1与an的关系,设an+1+x(n+1)+y=2(an+xn+y)进而可得方程组
解得x和y,代入an+1+x(n+1)+y=2(an+xn+y),可得an+1-(n+1)=2(an-n),进而可证明数列{an-n}为等比数列.
(2)把(1)中求得的an代入Cn,可得Cn=
=
[
-
],根据
=2-
<2可知Cn<2[
-
],进而可知C2+C3++Cn<2(
-
),原式得证.
(3)把bn代入bn+1=f(bn).可得bn+1+1=(bn+1)2,两边求对数化简得
=(
)2n-1≤(
)n,进而根据等比数列的求和公式可推断
+
+…+
≤
<2,又
+
++
≥
=
进而可证明原式.
|
(2)把(1)中求得的an代入Cn,可得Cn=
1 |
2n-1 |
2n+1-1 |
2n |
1 |
2n-1 |
1 |
2n+1-1 |
2n+1-1 |
2n |
1 |
2n |
1 |
2n-1 |
1 |
2n+1-1 |
1 |
3 |
1 |
2n+1-1 |
(3)把bn代入bn+1=f(bn).可得bn+1+1=(bn+1)2,两边求对数化简得
1 |
bn+1 |
1 |
3 |
1 |
3 |
1 |
1+b1 |
1 |
1+b2 |
1 |
1+bn |
1-(
| ||
2 |
1 |
b1+1 |
1 |
b2+1 |
1 |
bn+1 |
1 |
b1+1 |
1 |
3 |
解答:解:(1)f'(x)=2x+2?an+1=2an+2-n-1?an+1=2an-n+1
设an+1+x(n+1)+y=2(an+xn+y)?
?
?an+1-(n+1)=2(an-n),
∴数列{an-n}是首项为2,公比为2的等比数列,
∴an=2n+n(n∈N*).
(2)Cn=
=
[
-
]<2[
-
]
∴C2+C3++Cn<2[
-
+
-
++
-
]
=2(
-
)<
.
(3)bn+1=bn2+2bn?bn+1+1=(bn+1)2?log3(bn+1+1)=2log3(bn+1)
?
=(
)2n-1?
≤(
)n?
+
+
<
,
又∵
+
++
≥
=
∴原式得证.
设an+1+x(n+1)+y=2(an+xn+y)?
|
|
?an+1-(n+1)=2(an-n),
∴数列{an-n}是首项为2,公比为2的等比数列,
∴an=2n+n(n∈N*).
(2)Cn=
1 |
2n-1 |
2n+1-1 |
2n |
1 |
2n-1 |
1 |
2n+1-1 |
1 |
2n-1 |
1 |
2n+1-1 |
∴C2+C3++Cn<2[
1 |
22-1 |
1 |
23-1 |
1 |
23-1 |
1 |
24-1 |
1 |
2n-1 |
1 |
2n+1-1 |
=2(
1 |
3 |
1 |
2n+1-1 |
2 |
3 |
(3)bn+1=bn2+2bn?bn+1+1=(bn+1)2?log3(bn+1+1)=2log3(bn+1)
?
1 |
bn+1 |
1 |
3 |
1 |
bn+1 |
1 |
3 |
1 |
b1+1 |
1 |
b2+1 |
1 |
bn+1 |
1 |
2 |
又∵
1 |
b1+1 |
1 |
b2+1 |
1 |
bn+1 |
1 |
b1+1 |
1 |
3 |
∴原式得证.
点评:本题主要考查数列等比关系的确定.等比数列常与幂函数、对数函数、不等式一块考查,应注意联系这些函数的性质.
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