设数列{an}为单调递增的等差数列,a1=1,且a3,a6,a12依次成等比数列.(Ⅰ)求数列{an}的通项公式an;(Ⅱ)若bn=an•2an,求数列{bn}的前n项和Sn;(Ⅲ)若cn=2an(2

题目简介

设数列{an}为单调递增的等差数列,a1=1,且a3,a6,a12依次成等比数列.(Ⅰ)求数列{an}的通项公式an;(Ⅱ)若bn=an•2an,求数列{bn}的前n项和Sn;(Ⅲ)若cn=2an(2

题目详情

设数列{an}为单调递增的等差数列,a1=1,且a3,a6,a12依次成等比数列.
(Ⅰ)求数列{an}的通项公式an
(Ⅱ)若bn=an2an,求数列{bn}的前n项和Sn
(Ⅲ)若cn=
2an
(2an)2+3•2an+2
,求数列{cn}的前n项和Tn
题型:解答题难度:中档来源:绵阳二模

答案

(Ⅰ)∵数列{an}为单调递增的等差数列,a1=1,且a3,a6,a12依次成等比数列,
a12
a6
=
a6
a3
=
a12-a6
a6-a3
=class="stub"6d
3d
=2,
∴1+5d=2(1+2d),
解得d=1,
∴an=n.….(4分)
(Ⅱ)∵an=n,∴bn=an2an=n•2n
∴数列{bn}的前n项和Sn=1×2+2×22+3×23+…+n×2n,①
∴2Sn=1×22+2×23+3×24+…+n×2n+1,②
①-②,得-Sn=2+22+23+24+…+2n-n×2n+1
=
2×(1-2n)
1-2
-n×2n+1
=-(2-2n+1+n×2n+1),
∴Sn=2-2n+1+n×2n+1=(n-1)•2n+1+2.….(13分)
(Ⅲ)∵an=n,
cn=
2an
(2an)2+3•2an+2

=
2n
(2n)2+3×2n+2

=
2n
(2n+1)(2n+2)

=
2n-1
(2n+1)(2n-1+1)

=class="stub"1
2n-1+1
-class="stub"1
2n+1

∴数列{cn}的前n项和
Tn=(class="stub"1
20+1
-class="stub"1
21+1
)+(class="stub"1
21+1
-class="stub"1
22+1
)+…+(class="stub"1
2n-1+1
-class="stub"1
2n+1
)=class="stub"1
2
-class="stub"1
2n+1
.…(13分)

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