已知n是正整数,数列{an}的前n项和为Sn,数列{nan}的前n项和为Tn.对任何正整数n,等式Sn=-an+12(n-3)都成立.(I)求数列{an}的通项公式;(II)求Tn;(III)设An=

题目简介

已知n是正整数,数列{an}的前n项和为Sn,数列{nan}的前n项和为Tn.对任何正整数n,等式Sn=-an+12(n-3)都成立.(I)求数列{an}的通项公式;(II)求Tn;(III)设An=

题目详情

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

答案

(I) 当n=1时,由sn=-an+class="stub"1
2
(n-3)
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
=(-1)•(class="stub"1
2
)
n-1

an=class="stub"1
2
-class="stub"1
2n-1
,…7分

∴数列{an}的通项公式为an=class="stub"1
2
-class="stub"1
2n-1

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

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

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
2
)
n
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
…8分            
 (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
…10分
∵当n=2或n=3时,
-n2+5n-6
2
的值最大,最大值为0,
∴An-Bn≤0.
因此,当n是正整数时,An≤Bn.…12分

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