如图所示,流程图给出了无穷整数数列{an}满足的条件,a1∈N+,且当k=5时,输出的S=-59;当k=10时,输出的S=-1099.(1)试求数列{an}的通项公式an;(2)是否存在最小的正数M使

题目简介

如图所示,流程图给出了无穷整数数列{an}满足的条件,a1∈N+,且当k=5时,输出的S=-59;当k=10时,输出的S=-1099.(1)试求数列{an}的通项公式an;(2)是否存在最小的正数M使

题目详情

如图所示,流程图给出了无穷整数数列{an}满足的条件,a1∈N+,且当k=5时,输出的S=-
5
9
;当k=10时,输出的S=-
10
99

(1)试求数列{an}的通项公式an
(2)是否存在最小的正数M使得Tn≤M对一切正整数n都成立,若存在,求出M的值;若不存在,请说明理由.360优课网
题型:解答题难度:中档来源:不详

答案

(1)由题设知
class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
a5a6
=-class="stub"5
9
class="stub"1
a1a2
+class="stub"1
a2a3
+…+class="stub"1
a10a11
=-class="stub"10
99

又∵{an}是等差数列,设公差为d,
class="stub"1
d
(class="stub"1
a1
-class="stub"1
a6
)=-class="stub"5
9
class="stub"1
d
(class="stub"1
a1
-class="stub"1
a11
)=-class="stub"10
99
.
a1a6=-9
a1a11=-99.

两式相减得:a1(a11-a6)=-90,即a1d=-18
又∵a1d=a1(a1+5d)=a12-90,∴a12=81,
∴a1=9,a1=-9舍,∴d=-2,∴an=11-2n
(2)Tn=class="stub"9
20
+class="stub"7
21
+class="stub"5
22
+…+class="stub"11-2n
2n-1
.①
①式两边同乘class="stub"1
2
class="stub"1
2
Tn=class="stub"9
21
+class="stub"7
22
+…+class="stub"13-2n
2n-1
+class="stub"11-2n
2n
.②
②-①得(1-class="stub"1
2
)Tn=class="stub"9
20
+class="stub"-2
21
+class="stub"-2
22
…+class="stub"-2
2n-1
-class="stub"11-2n
2n

class="stub"1
2
Tn=9-2(class="stub"1
2
+class="stub"1
22
+…+class="stub"1
2n-1
)-class="stub"11-2n
2n
=9-2(1-class="stub"1
2n-1
)-class="stub"11-2n
2n

Tn=14+class="stub"2n-7
2n-1

又∵Tn+1-Tn=class="stub"2n-5
2n
-class="stub"2n-7
2n-1
=class="stub"9-2n
2n

当n≥5时,∵Tn+1-Tn<0;当n≤4时,
∵Tn+1-Tn>0∴当n=5时,Tn有最大值class="stub"227
16

∵Tn≤M恒成立,∴M≥class="stub"227
16

∴M的最小值为class="stub"227
16

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