设数列的前项的和,(Ⅰ)求首项与通项;(Ⅱ)设,,证明:.-高三数学

题目简介

设数列的前项的和,(Ⅰ)求首项与通项;(Ⅱ)设,,证明:.-高三数学

题目详情

设数列的前项的和
(Ⅰ)求首项与通项
(Ⅱ)设,证明:.
题型:解答题难度:偏易来源:不详

答案

(Ⅰ)a1=2,an=4n-2n, n="1,2,3," …,;(Ⅱ)同解析;
(Ⅰ)由 Sn=an-×2n+1+, n=1,2,3,… , ① 
得 a1=S1= a1-×4+所以a1=2.
再由①有 Sn-1=an-1-×2n+, n=2,3,4,…
将①和②相减得: an=Sn-Sn-1= (an-an-1)-×(2n+1-2n),n="2,3," …
整理得: an+2n=4(an-1+2n-1),n="2,3," … ,
因而数列{ an+2n}是首项为a1+2=4,公比为4的等比数列,
即 : an+2n=4×4n-1= 4n, n="1,2,3," …,
因而an=4n-2n, n="1,2,3," …,
(Ⅱ)将an=4n-2n代入①得:
Sn= ×(4n-2n)-×2n+1 + = ×(2n+1-1)(2n+1-2)
= ×(2n+1-1)(2n-1)   
Tn= = × = ×(- )
所以, = - )
= ×(- ) <

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