设数{an}的前n项和为Sn,且a1=1,an+1=2Sn+1,数列{bn}满足a1=b1,点P(bn,bn+1)在直线x-y+2=0上,n∈N*(1)求数列{an},{bn}的通项公式;(2)设cn

题目简介

设数{an}的前n项和为Sn,且a1=1,an+1=2Sn+1,数列{bn}满足a1=b1,点P(bn,bn+1)在直线x-y+2=0上,n∈N*(1)求数列{an},{bn}的通项公式;(2)设cn

题目详情

设数{an}的前n项和为Sn,且a1=1,an+1=2Sn+1,数列{bn}满足a1=b1,点P(bn,bn+1)在直线x-y+2=0上,n∈N*
(1)求数列{an},{bn}的通项公式;
(2)设cn=
bn
an
,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)由an+1=2Sn+1可得an=2Sn-1+1(n≥2),
两式相减得an+1-an=2an,
an+1=3an(n≥2).
又a2=2S1+1=3,
所以a2=3a1.
故{an}是首项为1,公比为3的等比数列.
所以an=3n-1.
由点P(bn,bn+1)在直线x-y+2=0上,所以bn+1-bn=2.
则数列{bn}是首项为1,公差为2的等差数列.
则bn=1+(n-1)•2=2n-1
(Ⅱ)因为 cn=
bn
an
=class="stub"2n-1
3n-1
,所以 Tn=class="stub"1
30
+class="stub"3
31
+class="stub"5
32
++class="stub"2n-1
3n-1

class="stub"1
3
Tn=class="stub"1
31
+class="stub"3
32
+class="stub"5
32
++class="stub"2n-3
3n-1
+class="stub"2n-1
3n

两式相减得:class="stub"2
3
Tn=1+class="stub"2
3
+class="stub"2
32
++class="stub"2
3n-1
-class="stub"2n-1
3n

所以 Tn=3-class="stub"1
2•3n-2
-class="stub"2n-1
2•3n-1
=3-class="stub"n+1
3n-1

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