燕影翩跹
幼苗
共回答了15个问题采纳率:86.7% 举报
解题思路:(I)n=1时,6a
1=a
12+3a
1+2,且a
1>1,解得a
1=2.n≥2时,6S
n=a
n2+3a
n+2,6S
n-1=a
n-12+3a
n-1+2,两式相减得(a
n+a
n-1)(a
n-a
n-1-3)=0由此能求出a
n.
(II)根据数列{b
n}满足
an(2bn−1)=1,可得
bn=log2,从而T
n=b
1+b
2+…+b
n=
log2(××…×),利用分析法证明.要证2T
n+1<log
2(a
n+3),即证
2log2(××…×)+1<log
2(a
n+3),即证
<1,构造函数
cn=,可得{c
n}是单调递减数列,即可证出结论.
(I)n=1时,6a1=a12+3a1+2,且a1>1,解得a1=2.
n≥2时,6Sn=an2+3an+2,6Sn-1=an-12+3an-1+2,两式相减得(an+an-1)(an-an-1-3)=0,
∵an+an-1>0,
∴an-an-1=3,
∴{an}为等差数列,
∵a1=2,
∴an=3n-1.
(II)证明:∵数列{bn}满足an(2bn−1)=1,
∴bn=log2
3n
3n−1
∴Tn=b1+b2+…+bn=log2(
3
2×
6
5×…×
3n
3n−1)
要证2Tn+1<log2(an+3),即证2log2(
3
2×
6
5×…×
3n
3n−1)+1<log2(an+3)
即证(
3
2×
6
5×…×
3n
3n−1)2<
3n+2
2
即证
2(
3
2×
6
5×…×
3n
3n−1)2
3n+2<1
令cn=
2(
3
2×
6
5×…×
3n
3n−1)2
3n+2,
∴
cn+1
cn=
9n2+18n+9
9n2+21n+10<1
∵cn>0,∴cn+1<cn,
∴{cn}是单调递减数列
∴cn≤c1=
2×(
3
2)2
3×1+2=
9
10<1
∴cn=
2(
3
2
点评:
本题考点: 数列与不等式的综合;等差数列的通项公式;数列的求和;数列递推式.
考点点评: 本题考查数列的综合应用,解题时要认真审题,仔细解答,注意挖掘题设中的隐含条件,合理地进行等价转化.
1年前
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