首页 > 设数列{an}满足a1=1，a2=2，an=13（an-1+2an-2）（n=3，4，…）．数列{bn}满足b1=1，bn（n=2，3，…）是非零整数，且对任意的正整数m和自然数k，都有-1≤bm+b

设数列{an}满足a1=1，a2=2，an=13（an-1+2an-2）（n=3，4，…）．数列{bn}满足b1=1，bn（n=2，3，…）是非零整数，且对任意的正整数m和自然数k，都有-1≤bm+b

题目简介

设数列{an}满足a1=1，a2=2，an=13（an-1+2an-2）（n=3，4，…）．数列{bn}满足b1=1，bn（n=2，3，…）是非零整数，且对任意的正整数m和自然数k，都有-1≤bm+b

题目详情

设数列{an}满足a1=1，a2=2，an=
1
3
（an-1+2an-2）（n=3，4，…）．数列{bn}满足b1=1，bn（n=2，3，…）是非零整数，且对任意的正整数m和自然数k，都有-1≤bm+bm+1+…+bm+k≤1．
（1）求数列{an}和{bn}的通项公式；
（2）记cn=nanbn（n=1，2，…），求数列{cn}的前n项和Sn．
题型：解答题难度：中档来源：广东

答案

（1）由an=class="stub"1
3
(an-1-an-2)得an-an-1=-class="stub"2
3
(an-1-an-2)（n≥3）
又a2-a1=1≠0，
∴数列{an+1-an}是首项为1公比为-class="stub"2
3
的等比数列，an+1-an=(-class="stub"2
3
)n-1
an=a1+（a2-a1）+（a3-a2）+（a4-a3）++（an-an-1）
=1+1+(-class="stub"2
3
)+(-class="stub"2
3
)2++(-class="stub"2
3
)n-2
=1+
1-(-class="stub"2
3
)n-1
1+class="stub"2
3
=class="stub"8
5
-class="stub"3
5
(-class="stub"2
3
)n-1，
当n为奇数时当n为偶数时
由
-1≤b1+b2≤1
-1≤b2≤1
b2∈Z，b2≠0

得b2=-1，
由
-1≤b2+b3≤1
-1≤b3≤1
b3∈Z，b3≠0

得b3=1，
同理可得当n为偶数时，bn=-1；当n为奇数时，bn=1；
因此bn=
1，n为奇数
-1，n为偶数
．
（2）cn=nanbn=
class="stub"8
5
n-class="stub"3
5
n(class="stub"2
3
)n-1
-class="stub"8
5
n-class="stub"3
5
n(class="stub"2
3
)n-1

Sn=c1+c2+c3+c4++cn
当n为奇数时，Sn=(class="stub"8
5
-2×class="stub"8
5
+3×class="stub"8
5
-4×class="stub"8
5
++class="stub"8
5
n)-class="stub"3
5
[1×(class="stub"2
3
)0+2×(class="stub"2
3
)1+3×(class="stub"2
3
)2+4×(class="stub"2
3
)3++n(class="stub"2
3
)n-1]=
4(n+1)
5
-class="stub"3
5
[1×(class="stub"2
3
)0+2×(class="stub"2
3
)1+3×(class="stub"2
3
)2+4×(class="stub"2
3
)3++n(class="stub"2
3
)n-1]
当n为偶数时
Sn=(class="stub"8
5
-2×class="stub"8
5
+3×class="stub"8
5
-4×class="stub"8
5
+-class="stub"8
5
n)-class="stub"3
5
[1×(class="stub"2
3
)0+2×(class="stub"2
3
)1+3×(class="stub"2
3
)2+4×(class="stub"2
3
)3++n(class="stub"2
3
)n-1]=-class="stub"4n
5
-class="stub"3
5
[1×(class="stub"2
3
)0+2×(class="stub"2
3
)1+3×(class="stub"2
3
)2+4×(class="stub"2
3
)3++n(class="stub"2
3
)n-1]
令Tn=1×(class="stub"2
3
)0+2×(class="stub"2
3
)1+3×(class="stub"2
3
)2+4×(class="stub"2
3
)3++n(class="stub"2
3
)n-1①
①×class="stub"2
3
得：class="stub"2
3
Tn=1×(class="stub"2
3
)1+2×(class="stub"2
3
)2+3×(class="stub"2
3
)3+4×(class="stub"2
3
)4++n(class="stub"2
3
)n②
①-②得：class="stub"1
3
Tn=1+(class="stub"2
3
)1+(class="stub"2
3
)2+(class="stub"2
3
)3+(class="stub"2
3
)4++(class="stub"2
3
)n-1-n(class="stub"2
3
)n=
1-(class="stub"2
3
)n
1-class="stub"2
3
-n(class="stub"2
3
)n=3-(3+n)(class="stub"2
3
)n
∴Tn=9-(9+3n)(class="stub"2
3
)n
当n为奇数时当n为偶数时
因此Sn=
class="stub"4n-23
5
+
9(n+3)
5
(class="stub"2
3
)n
-class="stub"4n+27
5
+
9(n+3)
5
(class="stub"2
3
)n

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