已知数列{an}的前n项和为Sn.且满足Sn=2an-1(n∈N+)(I)求证:数列{an}是等比数列;(II)数列{bn}满足bn+1.=an+bnn∈N+.且b1=3.若不等式log2(bn-2)

题目简介

已知数列{an}的前n项和为Sn.且满足Sn=2an-1(n∈N+)(I)求证:数列{an}是等比数列;(II)数列{bn}满足bn+1.=an+bnn∈N+.且b1=3.若不等式log2(bn-2)

题目详情

已知数列{an}的前n项和为Sn.且满足Sn=2an-1(n∈N+
(I)求证:数列{an}是等比数列;
(II)数列{bn}满足bn+1.=an+bnn∈N+.且b1=3.若不等式log2(bn-2)
3
16
n2+t
对任意n∈N+恒成立,求实数t的取值范围.
题型:解答题难度:中档来源:不详

答案

( I)证明:依题意可得Sn+1=2an+1-1…①,Sn=2an-1…②
①-②,得an+1=2an+1-2an
化简得
an+1
an
=2(n∈N*)

∵a1=2a1-1,
∴a1=1
∴数列{an}是以1为首项,公比为2的等比数列.
(II)由(Ⅰ)可知an=2n-1,因为bn+1=an+bn,n∈N+.且b1=3,
所以bn=an-1+bn-1=an-1+an-2+bn-2=…=an-1+an-2+…+a1+b1
=2n-2+2n-3+…+1+3=2n-1+2,
因为不等式log2(bn-2)<class="stub"3
16
n2+t
对任意n∈N+恒成立,
所以log2(2n-1+2-2)<class="stub"3
16
n2+t

即t>-class="stub"3
16
n2+n-1
,对任意n∈N+恒成立,
因为-class="stub"3
16
n2+n-1≤class="stub"5
16
,且n=3时-class="stub"3
16
n2+n-1
取得最大值class="stub"5
16

所以t>class="stub"5
16

所以实数t的取值范围:(class="stub"5
16
,+∞)

更多内容推荐