问题
解答题
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与
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答案
(Ⅰ)在等差数列中,设公差为d≠0,
由题意
,∴a1a5= a 22 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,
=1,此时Tn=3×1-1 1+1
.3n-1 n+1
当n≥2时,Tn=1+4(
+1 22
+…+1 23
)1 2n
=1+
=3-4×
(1-1 22
)1 2n-1 1- 1 2
.1 2n-2
又2n=(1+1)n=
+C 0n
+…+C 1n
>n+1,C nn
∴
=1 2n-2
<4 2n
,3-4 n+1
>3-1 2n-2
=4 n+1
.3n-1 n+1
∴当n=1时,Tn=
,当n≥2时,Tn>3n-1 n+1
.3n-1 n+1