已知数列{an}满足11-an+1-11-an=1,且a1=0.(1)求数列{an}的通项公式;(2)设bn=n•2nan,求数列{bn}的前n项和Sn;(3)设cn=1-an+1n,记Tn=nk=1

题目简介

已知数列{an}满足11-an+1-11-an=1,且a1=0.(1)求数列{an}的通项公式;(2)设bn=n•2nan,求数列{bn}的前n项和Sn;(3)设cn=1-an+1n,记Tn=nk=1

题目详情

已知数列{an}满足
1
1-an+1
-
1
1-an
=1
,且a1=0.
(1)求数列{an}的通项公式;
(2)设bn=n•2nan,求数列{bn}的前n项和Sn
(3)设cn=
1-
an+1
n
,记Tn=
n


k=1
ck
,证明:Tn<1.
题型:解答题难度:中档来源:不详

答案

(1)∵a1=0,∴class="stub"1
1-a1
=1

class="stub"1
1-an+1
-class="stub"1
1-an
=1

∴数列{class="stub"1
1-an
}是以1为首项,1为公差的等差数列
class="stub"1
1-an
=n,∴an=class="stub"n-1
n

(2)bn=n•2nan=(n-1)•2n,
∴Sn=1•22+2•23+…+(n-1)•2n,
∴2Sn=1•23+2•24+…+(n-2)•2n+(n-1)•2n+1,
两式相减可得-Sn=1•22+1•23+…+1•2n-(n-1)•2n+1,
∴Sn=4+(n-2)•2n+1;
(3)证明:cn=
1-
an+1
n
=class="stub"1
n
-class="stub"1
n+1

∴Tn=
n


k=1
ck
=1-class="stub"1
2
+class="stub"1
2
-class="stub"1
3
+…+class="stub"1
n
-class="stub"1
n+1
=1-class="stub"1
n+1
<1,
∴Tn<1.

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