已知等差数列{an}的公差不为零,且a3=5,a1,a2.a5成等比数列(I)求数列{an}的通项公式:(II)若数列{bn}满足b1+2b2+4b3+…+2n-1bn=an且数列{bn}的前n项和T

题目简介

已知等差数列{an}的公差不为零,且a3=5,a1,a2.a5成等比数列(I)求数列{an}的通项公式:(II)若数列{bn}满足b1+2b2+4b3+…+2n-1bn=an且数列{bn}的前n项和T

题目详情

已知等差数列{an}的公差不为零,且a3=5,a1,a2.a5 成等比数列
(I)求数列{an}的通项公式:
(II)若数列{bn}满足b1+2b2+4b3+…+2n-1bn=an且数列{bn}的前n项和Tn 试比较Tn
3n-1
n+1
的大小.
题型:解答题难度:中档来源:嘉兴一模

答案

(Ⅰ)在等差数列中,设公差为d≠0,
由题意
a1a5=
a22
a3=5
,∴
a1(a1+4d)=(a1+d)2
a1+2d=5

解得
a1=1
d=2

∴an=a1+(n-1)d=1+2(n-1)=2n-1.
(Ⅱ)∵b1+2b2+4b3+…+2n-1bn=an,①
b1+2b2+4b3+…+2n-1bn+2nbn=an+1,②
②-①得2nbn+1=2,∴bn+1=21-n
当n=1时,b1=a1=1,∴bn=
22-n,当m≥2时
1,当n=1时

当n=1时,T1=a1=1,class="stub"3×1-1
1+1
=1
,此时Tn=class="stub"3n-1
n+1

当n≥2时,Tn=1+4(class="stub"1
22
+class="stub"1
23
+…+class="stub"1
2n
)

=1+
4×class="stub"1
22
(1-class="stub"1
2n-1
)
1-class="stub"1
2
=3-class="stub"1
2n-2

2n=(1+1)n=
C0n
+
C1n
+…+
Cnn
>n+1,
class="stub"1
2n-2
=class="stub"4
2n
<class="stub"4
n+1
3-class="stub"1
2n-2
>3-class="stub"4
n+1
=class="stub"3n-1
n+1

∴当n=1时,Tn=class="stub"3n-1
n+1
,当n≥2时,Tn>class="stub"3n-1
n+1

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