设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)求数列{an}的通项公式;(2)令cn=(-1)n+1logann+12,数列{cn}的前n项和为Tn,求证:当n∈N*且

题目简介

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).(1)求数列{an}的通项公式;(2)令cn=(-1)n+1logann+12,数列{cn}的前n项和为Tn,求证:当n∈N*且

题目详情

设数列{an}的前n项和为Sn,已知Sn=2an-2n+1(n∈N*).
(1)求数列{an}的通项公式;
(2)令cn=(-1)n+1log
an
n+1
2
,数列{cn}的前n项和为Tn,求证:当n∈N*且n≥2时,T2n
2
2
题型:解答题难度:中档来源:不详

答案

(1)由Sn=2an-2n+1,得Sn-1=2an-1-2n(n≥2).
两式相减,得an=2an-2an-1-2n,即an-2an-1=2n(n≥2).
于是
an
2n
-
an-1
2n-1
=1
,所以数列{
an
2n
}
是公差为1的等差数列.(5分)
又S1=2a1-22,所以a1=4.
所以
an
2n
=2+(n-1)=n+1
,故an=(n+1)•2n.(6分)
(2)因为cn=(-1)n+1•class="stub"1
n
,则当n≥2时,T2n=1-class="stub"1
2
+class="stub"1
3
-class="stub"1
4
++class="stub"1
2n-1
-class="stub"1
2n
=(1+class="stub"1
2
+class="stub"1
3
++class="stub"1
2n
)-2(class="stub"1
2
+class="stub"1
4
++class="stub"1
2n
)
=class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
2n
.(9分)

下面证class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
2n
2
2

g(x)=ln(x+1)-class="stub"x
x+1
(x>0)
,则g′(x)=class="stub"1
x+1
-class="stub"1
(x+1)2
=class="stub"x
(x+1)2
>0

∴g(x)在(0,+∞)时单调递增,g(x)>g(0)=0,即当x>0时,ln(x+1)>class="stub"x
x+1

x=class="stub"1
n
lnclass="stub"n+1
n
>class="stub"1
n+1
⇒ln(n+1)-lnn>class="stub"1
n+1
ln(n+2)-ln(n+1)>class="stub"1
n+2

ln(n+3)-ln(n+2)>class="stub"1
n+3
,,ln(2n)-ln(2n-1)>class="stub"1
2n

以上n个式相加,即有ln(2n)-lnn>class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
2n

class="stub"1
n+1
+class="stub"1
n+2
++class="stub"1
2n
<ln(2n)-lnn=ln2<
2
2
(14分)

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