已知n是正整数,数列{an}的前n项和为Sn,且满足Sn=-an+12(n-3),数列(nan)的前n项和为Tn.(1)求数列{an}的通项公式;(2)求Tn;(3)设An=2Tn,Bn=(2n+4)

题目简介

已知n是正整数,数列{an}的前n项和为Sn,且满足Sn=-an+12(n-3),数列(nan)的前n项和为Tn.(1)求数列{an}的通项公式;(2)求Tn;(3)设An=2Tn,Bn=(2n+4)

题目详情

已知n是正整数,数列{an}的前n项和为Sn,且满足Sn=-an+
1
2
(n-3),数列(nan)的前n项和为Tn
(1)求数列{an}的通项公式;
(2)求Tn
(3)设An=2Tn,Bn=(2n+4)Sn+3,试比较An与Bn的大小.
题型:解答题难度:中档来源:不详

答案

(1)当n=1时,由已知可得,S1=a1=-a1+class="stub"1
2
(1-3)

解得a1=-class="stub"1
2
…2分
当n≥2时,an=Sn-Sn-1=-an+class="stub"1
2
(n-3)-[-an-1+class="stub"1
2
(n-4)]
解得 an=class="stub"1
2
an-1+class="stub"1
4
,即an-class="stub"1
2
class="stub"1
2
(an-1-class="stub"1
2
)

因此,数列{an-class="stub"1
2
}是首项为-1,公比为 class="stub"1
2
的等比数列
an-class="stub"1
2
=-(class="stub"1
2
)n-1

∴an=class="stub"1
2
-class="stub"1
2n-1

(II)∵nan=class="stub"n
2
-class="stub"n
2n-1

∴Tn=(1+2+3+…+n)-(1+2×class="stub"1
2
+3×class="stub"1
22
+…+n×class="stub"1
2n-1
)…6分
令Un=1+2×class="stub"1
2
+3×class="stub"1
22
+…+n×class="stub"1
2n-1

class="stub"1
2
Un=class="stub"1
2
+2×class="stub"1
22
+3×class="stub"1
23
+…+n×class="stub"1
2n

上面两式相减:class="stub"1
2
Un=1+class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n-1
-n×class="stub"1
2n

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

即Un=4-class="stub"n+2
2n-1

∴Tn=
n(n+1)
4
-4+class="stub"n+2
2n-1
=
n2+n-16
4
+class="stub"n+2
2n-1

(III)∵Sn=-an+class="stub"n-3
2

=-class="stub"1
2
+class="stub"1
2n-1
+class="stub"n-3
2

=class="stub"n-4
2
+class="stub"1
2n-1

An-Bn=
n2+n-16
2
+class="stub"n+2
2n-2
-
(2n+4)(n-4)
2
-class="stub"n+2
2n-2
-3

=
-n2+5n-6
2

∵当n=2或n=3时,
-n2+5n-6
2
的值最大,最大值为0,
∴An-Bn≤0.
∴An≤Bn.

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