.已知各项均为正数的数列{an}满足a1=1,an+1+an•an+1-an=0.(Ⅰ)求证:数列{1an}是等差数列;(Ⅱ)求数列{2nan}前n项和Sn.-数学

题目简介

.已知各项均为正数的数列{an}满足a1=1,an+1+an•an+1-an=0.(Ⅰ)求证:数列{1an}是等差数列;(Ⅱ)求数列{2nan}前n项和Sn.-数学

题目详情

.已知各项均为正数的数列{an}满足a1=1,an+1+an•an+1-an=0.
(Ⅰ)求证:数列{
1
an
}
是等差数列;
(Ⅱ)求数列{
2n
an
}
前n项和Sn
题型:解答题难度:中档来源:大连一模

答案

(Ⅰ)∵an+1+an•an+1-an=0,
an+1+anan+1-an
anan+1
=0,
class="stub"1
an+1
-class="stub"1
an
=1,(3分)
class="stub"1
a1
=1,
∴数列{class="stub"1
an
}是以1为首项,1为公差的等差数列.(4分)
class="stub"1
an
=1+(n-1)×1=n,an=class="stub"1
n
.(6分)
(Ⅱ)由(Ⅰ)知
2n
an
=n•2n.
Sn=1×21+2×22+…+n×2n.①
2Sn=1×22+2×23+…+n×2n+1.②(9分)
由①-②得-Sn=21+22+…+2n-n×2n+1.
∴Sn=(n-1)2n+1+2.(12分)

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