设公差不为0的等差数列{an}的首项为1,且a2,a5,a14构成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)若数列{bn}满足b1a1+b2a2+…+bnan=1-12n,n∈N*,求{bn}的

题目简介

设公差不为0的等差数列{an}的首项为1,且a2,a5,a14构成等比数列.(Ⅰ)求数列{an}的通项公式;(Ⅱ)若数列{bn}满足b1a1+b2a2+…+bnan=1-12n,n∈N*,求{bn}的

题目详情

设公差不为0的等差数列{an}的首项为1,且a2,a5,a14构成等比数列.
(Ⅰ)求数列{an}的通项公式;
(Ⅱ)若数列{bn}满足
b1
a1
+
b2
a2
+…+
bn
an
=1-
1
2n
,n∈N*,求{bn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)设等差数列{an}的公差为d(d≠0),
∵a2,a5,a14构成等比数列,
a52=a2a14,即(1+4d)2=(1+d)(1+13d),
解得d=0(舍去),或d=2.
∴an=1+(n-1)×2=2n-1.
(Ⅱ)由已知,
b1
a1
+
b2
a2
+…+
bn
an
=1-class="stub"1
2n
,n∈N*,
当n=1时,class="stub"b1
a1
=class="stub"1
2

当n≥2时,class="stub"bn
an
=1-class="stub"1
2n
-(1-class="stub"1
2n-1
)=class="stub"1
2n

class="stub"bn
an
=class="stub"1
2n
,n∈N*.
由(Ⅰ),知an=2n-1,n∈N*,
∴bn=class="stub"2n-1
2n
,n∈N*.
又Tn=class="stub"1
2
+class="stub"3
22
+class="stub"5
23
+…+class="stub"2n-1
2n

class="stub"1
2
Tn=class="stub"1
22
+class="stub"3
23
+…+class="stub"2n-3
2n
+class="stub"2n-1
2n+1

两式相减,得class="stub"1
2
Tn=class="stub"1
2
+(class="stub"2
22
+class="stub"2
23
+…+class="stub"2
2n
)-class="stub"2n-1
2n+1
=class="stub"3
2
-class="stub"1
2n-1
-class="stub"2n-1
2n+1

∴Tn=3-class="stub"2n+3
2n

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