已知数列{an}的前n项和是Sn,且2Sn=2-an.(Ⅰ)求数列{an}的通项公式;(Ⅱ)记bn=an+n,求数列{bn}的前n项和Tn.-数学

题目简介

已知数列{an}的前n项和是Sn,且2Sn=2-an.(Ⅰ)求数列{an}的通项公式;(Ⅱ)记bn=an+n,求数列{bn}的前n项和Tn.-数学

题目详情

已知数列{an}的前n项和是Sn,且2Sn=2-an
(Ⅰ)求数列{an}的通项公式;
(Ⅱ) 记bn=an+n,求数列{bn}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,2S1=2-a1,2a1=2-a1,∴a1=class="stub"2
3

当n≥2时,
2Sn=2-an
2Sn-1=2-an-1

两式相减得2an=an-1-an(n≥2),
即3an=an-1(n≥2),又an-1≠0∴
an
an-1
=class="stub"1
3
(n≥2),
∴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=2•(class="stub"1
3
)n

(Ⅱ)由(Ⅰ)知,bn=2•(class="stub"1
3
)n+n

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

=1-(class="stub"1
3
)n+
n2+n
2

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