数列{an}的前n项和是Sn,若数列{an}的各项按如下规则排列:12,13,23,14,24,34,15,25,35,45,16,…,若存在整数k,使Sk<10,Sk+1≥10,则ak=______

题目简介

数列{an}的前n项和是Sn,若数列{an}的各项按如下规则排列:12,13,23,14,24,34,15,25,35,45,16,…,若存在整数k,使Sk<10,Sk+1≥10,则ak=______

题目详情

数列{an}的前n项和是Sn,若数列{an}的各项按如下规则排列:
1
2
1
3
2
3
1
4
2
4
3
4
1
5
2
5
3
5
4
5
1
6
,…,若存在整数k,使Sk<10,Sk+1≥10,则ak=______.
题型:填空题难度:中档来源:深圳二模

答案

把原数列划分成class="stub"1
2
;class="stub"1
3
,class="stub"2
3
;class="stub"1
4
,class="stub"2
4
,class="stub"3
4
;class="stub"1
5
,class="stub"2
5
,class="stub"3
5
,class="stub"4
5
;class="stub"1
6
,…

发现他们的个数是1,2,3,4,5…
构建新数列bn,则bn=class="stub"1
2
n
是个等差数列,记bn的前n项和为Tn,
利用等差数列的和知道T5=class="stub"15
2
T6=class="stub"21
2

所以ak定在class="stub"1
7
,class="stub"2
7
,class="stub"3
7
,…,class="stub"6
7

又因为Sk<10,Sk+1≥10,而T5+class="stub"1
7
+class="stub"2
7
+class="stub"3
7
+class="stub"4
7
+class="stub"5
7
=9+class="stub"9
14
<10

T5+class="stub"1
7
+class="stub"2
7
+class="stub"3
7
+class="stub"4
7
+class="stub"5
7
+class="stub"6
7
=10+class="stub"1
2
>10

所以ak=class="stub"5
7

故答案为:class="stub"5
7

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