已知各项均为正数的数列{an}的前n项和满足S1>1,且6Sn=(an+1)(an+2),n∈N*.(1)求{an}的通项公式;(2)设数列{bn}满足an(2bn-1)=1,并记Tn为{bn}的前n

题目简介

已知各项均为正数的数列{an}的前n项和满足S1>1,且6Sn=(an+1)(an+2),n∈N*.(1)求{an}的通项公式;(2)设数列{bn}满足an(2bn-1)=1,并记Tn为{bn}的前n

题目详情

已知各项均为正数的数列{an}的前n项和满足S1>1,且6Sn=(an+1)(an+2),n∈N*
(1)求{an}的通项公式;
(2)设数列{bn}满足an(2bn-1)=1,并记Tn为{bn}的前n项和,求证:3Tn+1>log2(an+3),n∈N*
题型:解答题难度:中档来源:重庆

答案

(1)由a1=S1=class="stub"1
6
(a1+1)(a1+2)
,解得a1=1或a1=2,由假设a1=S1>1,因此a1=2,
又由an+1=Sn+1-Sn=class="stub"1
6
(an+1+1)(an+1+2)-class="stub"1
6
(an+1)(an+2)

得(an+1+an)(an+1-an-3)=0,
即an+1-an-3=0或an+1=-an,因an>0,故an+1=-an不成立,舍去
因此an+1-an=3,从而{an}是公差为3,首项为2的等差数列,
故{an}的通项为an=3n-1
证明:由an(2bn-1)=1可解得bn=log2(1+class="stub"1
a2
)=log2class="stub"3n
3n-1

从而Tn=b1+b2++bn=log2(class="stub"3
2
•class="stub"6
5
••class="stub"3n
3n-1
)

因此3Tn+1-log2(an+3)=log2(class="stub"3
2
•class="stub"6
5
••class="stub"3n
3n-1
)3•class="stub"2
3n+2

f(n)=(class="stub"3
2
•class="stub"6
5
••class="stub"3n
3n-1
)3•class="stub"2
3n+2
,则
f(n+1)
f(n)
=class="stub"3n+2
3n+5
(class="stub"3n+3
3n+2
)
3
=
(3n+3)3
(3n+5)(3n+2)2

因(3n+3)3-(3n+5)(3n+2)2=9n+7>0,故f(n+1)>f(n)
特别地f(n)≥f(1)=class="stub"27
20
>1
,从而3Tn+1-log2(an+3)=log2f(n)>0、
即3Tn+1>log2(an+3)

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