数列{an}中,a1=1,a2=4,an=2n-1+λn2+μn,(n∈N*).(Ⅰ)求λ、μ的值;(Ⅱ)设数列{bn}满足:bn=1an+2n-2n-1,求数列{bn}的前n项和Sn.-数学

题目简介

数列{an}中,a1=1,a2=4,an=2n-1+λn2+μn,(n∈N*).(Ⅰ)求λ、μ的值;(Ⅱ)设数列{bn}满足:bn=1an+2n-2n-1,求数列{bn}的前n项和Sn.-数学

题目详情

数列{an}中,a1=1,a2=4,an=2n-1+λn2+μn,(n∈N*).
(Ⅰ)求λ、μ的值;
(Ⅱ)设数列{bn}满足:bn=
1
an+2n-2n-1
,求数列{bn}的前n项和Sn
题型:解答题难度:中档来源:湛江二模

答案

(Ⅰ)根据题意,得
4=2+4λ+2μ
1=1+λ+μ
(3分)
解得
λ=1
μ=-1
(6分)
(Ⅱ)由(Ⅰ)an=2n-1+n2-n
bn=class="stub"1
2n-1+n2-n-2n-1+2n
=class="stub"1
n2+n
=class="stub"1
n
-class="stub"1
n+1
(10分)
Sn=(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
=class="stub"n
n+1
(14分)

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