已知数列{an}的前n项和为Sn,且Sn=1-12an(n∈N*)(Ⅰ)求数列{an}的通项公式;(Ⅱ)已知数列{bn}的通项公式bn=2n-1,记cn=anbn,求数列{cn}的前n项和Tn.-数学

题目简介

已知数列{an}的前n项和为Sn,且Sn=1-12an(n∈N*)(Ⅰ)求数列{an}的通项公式;(Ⅱ)已知数列{bn}的通项公式bn=2n-1,记cn=anbn,求数列{cn}的前n项和Tn.-数学

题目详情

已知数列{an}的前n项和为Sn,且Sn=1-
1
2
an(n∈N*)

(Ⅰ)求数列{an}的通项公式;
(Ⅱ)已知数列{bn}的通项公式bn=2n-1,记cn=anbn,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)当n=1时,a1=1-class="stub"1
2
a1
,∴a1=class="stub"2
3

当n≥2时,an=Sn-Sn-1=1-class="stub"1
2
an-1+class="stub"1
2
an-1

class="stub"3
2
an=class="stub"1
2
an-1
,∴
an
an-1
=class="stub"1
3

∴数列{an}是以class="stub"2
3
为首项,class="stub"1
3
为公比的等比数列
an=class="stub"2
3
×(class="stub"1
3
)n-1=class="stub"2
3n
.…(6分)
(Ⅱ)∵cn=(2n-1)•class="stub"2
3n
,∴Tn=2[1×class="stub"1
3
+3×class="stub"1
32
+…+(2n-1)×class="stub"1
3n
]
.①
class="stub"1
3
Tn=2[1×class="stub"1
32
+3×class="stub"1
33
+…+(2n-1)×class="stub"1
3n+1
]
.②
①-②,得class="stub"2
3
Tn=2[class="stub"1
3
+class="stub"2
32
+…+class="stub"2
3n
-(2n-1)×class="stub"1
3n+1
]

class="stub"2
3
Tn=2[class="stub"1
3
+2•
class="stub"1
9
(1-class="stub"1
3n-1
)
1-class="stub"1
3
-(2n-1)•class="stub"1
3n+1
]

Tn=2-class="stub"2n+2
3n
(n∈N*).…(12分)

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