已知数列{an}的前n项和为Sn,且Sn=2an-2,(n=1,2,3,…);数列{bn}中,b1=1,点p(bn,bn+1)在直线x-y+2=0上.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)设

题目简介

已知数列{an}的前n项和为Sn,且Sn=2an-2,(n=1,2,3,…);数列{bn}中,b1=1,点p(bn,bn+1)在直线x-y+2=0上.(Ⅰ)求数列{an}和{bn}的通项公式;(Ⅱ)设

题目详情

已知数列{an} 的前n项和为Sn,且Sn=2an-2,(n=1,2,3,…);数列 {bn}中,b1=1,点p(bn,bn+1)在直线x-y+2=0上.
(Ⅰ)求数列{an} 和 {bn}的通项公式;
(Ⅱ)设数列{
bn+1
2
}的前n和为Sn,求
1
S1
+
1
S2
+…+
1
Sn

(Ⅲ)设数列{cn}的前n项和为Tn,且cn=an•bn,求Tn
题型:解答题难度:中档来源:不详

答案

(Ⅰ)∵Sn=2an-2,
∴当n≥2时,an=Sn-Sn-1=(2an-2)-(2an-1-2),…(1分)
即an=2an-1,
∴数列{an}是等比数列.
∵a1=S1=2a1-2,∴a1=2
∴an=2n.                           …(3分)
∵点P(bn,bn+1)在直线x-y+2=0上,
∴bn+1-bn=2,
即数列{bn}是等差数列,
又b1=1,∴bn=2n-1.…(5分)
(Ⅱ)由题意可得
bn+1
2
=n
,∴Sn=
n(n+1)
2
,…(6分)
class="stub"1
Sn
=2(class="stub"1
n
-class="stub"1
n+1
),…(7分)
class="stub"1
S1
+class="stub"1
S2
+…+class="stub"1
Sn
=2[(1-class="stub"1
2
)+(class="stub"1
2
-class="stub"1
3
)+…+(class="stub"1
n
-class="stub"1
n+1
)]=class="stub"2n
n+1
.…(9分)
(Ⅲ)∵cn=anbn=(2n-1)•2n…(10分)
Tn=1×2+3×22+5×23+…+(2n-3)2n-1+(2n-1)2n2Tn=1×22+3×23+5×24+…+(2n-3)2n+(2n-1)2n+1…(11分)
两式相减得:-Tn=2+2×(22+23+24+…+2n)-(2n-1)2n+1
=-6-(2n-3)2n+1…(13分)
Tn=6+(2n-3)2n+1…(14分)

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