已知数列{an}的前n项和Sn=n+n22k-1(n∈N*,k是与n无关的正整数).(1)求数列{an}的通项公式,并证明数列{an}是等差数列;(2)设数列{an}满足不等式:|a1-1|+|a2-

题目简介

已知数列{an}的前n项和Sn=n+n22k-1(n∈N*,k是与n无关的正整数).(1)求数列{an}的通项公式,并证明数列{an}是等差数列;(2)设数列{an}满足不等式:|a1-1|+|a2-

题目详情

已知数列{an}的前n项和Sn=
n+n2
2k-1
(n∈N*,k是与n无关的正整数).
(1)求数列{an}的通项公式,并证明数列{an}是等差数列;
(2)设数列{an}满足不等式:|a1-1|+|a2-1|+…|a2k-1-1|+|a2k-1|≤6,求所有这样的k的值.
题型:解答题难度:中档来源:不详

答案

(1)∵Sn=
n+n2
2k-1
(k是与n无关的正整数),
∴a1=class="stub"2
2k-1

当n≥2时,an=Sn-Sn-1=class="stub"1
2k-1
[(n2+n)-((n-1)2+(n-1))]=class="stub"2n
2k-1

当n=1时,a1=class="stub"2
2k-1
也适合上式,
∴an=class="stub"2n
2k-1

∴an+1-an=class="stub"1
2k-1
[2(n+1)-2n]=class="stub"2
2k-1
为定值,
∴数列{an}是等差数列;
(2)∵an=class="stub"2n
2k-1

∴ak=class="stub"2k
2k-1
=1+class="stub"1
2k-1

∴ak-1=class="stub"1
2k-1

又数列{an}的公差d=class="stub"2
2k-1
>0,故数列{an}为递增数列,
∴ak+1-1>class="stub"1
2k-1

ak+2-1>class="stub"1
2k-1
,…,
ak+k-1>class="stub"1
2k-1

∴|ak-1|+|ak+1-1|+…+|ak+k-1|=ak-class="stub"1
2k-1
+ak+1-+class="stub"1
2k-1
…+ak+k-class="stub"1
2k-1
>k+1,
(k+1)×2k
2k-1
+
(k+1)•k
2
class="stub"2
2k-1
>k+1+class="stub"k+1
2k-1

要使|a1-1|+|a2-1|+…|a2k-1-1|+|a2k-1|≤6,
需k+1<5(k∈N*),即1≤k≤4(k∈N*),
①当k=1时,a1=class="stub"2
2k-1
=2,d=class="stub"2
2k-1
=2,
∴an=2+(n-1)×2=2n,
∴|a1-1|+|a2-1|+…|a2k-1-1|+|a2k-1|=|a1-1|+|a2-1|=|2-1|+|4-1|=4≤6,即k=1时符合题意;
②当k=2时,a1=class="stub"2
2k-1
=class="stub"2
3
,d=class="stub"2
2k-1
=class="stub"2
3

同理可求an=class="stub"2n
3

∴|a1-1|+|a2-1|+…|a2k-1-1|+|a2k-1|=|a1-1|+|a2-1|+…+|a4-1|=(1-class="stub"2
3
)+(class="stub"4
3
-1)+(2-1)+(class="stub"8
3
-1)=class="stub"10
3
<6,故k=2时符合题意;
③当k=3时,同理可求an=class="stub"2
5
n,
|a1-1|+|a2-1|+…+|a6-1|=class="stub"3
5
+(1-class="stub"4
5
)+(class="stub"6
5
-1)+(class="stub"8
5
-1)+(2-1)+(class="stub"12
5
-1)=4<6,故k=3时符合题意;
④当k=4时,同理可求an=class="stub"2
7
n,
|a1-1|+|a2-1|+…+|a8-1|=class="stub"5
7
+class="stub"3
7
+class="stub"1
7
+class="stub"1
7
+(class="stub"10
7
-1)+(class="stub"12
7
-1)+(class="stub"14
7
-1)+(class="stub"16
7
-1)=class="stub"34
7
<6.故k=4时符合题意;
综上所述,存在k=1,2,3,4使|a1-1|+|a2-1|+…|a2k-1-1|+|a2k-1|≤6成立.

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