设数列{an}满足a1=1,a2=2,an=13(an-1+2an-2)(n∈N*且n≥3),bn=1n为奇数-1n为偶数(1)求an;(2)若cn=nanbn,n∈N*,求{cn}的前n项和Sn.-

题目简介

设数列{an}满足a1=1,a2=2,an=13(an-1+2an-2)(n∈N*且n≥3),bn=1n为奇数-1n为偶数(1)求an;(2)若cn=nanbn,n∈N*,求{cn}的前n项和Sn.-

题目详情

设数列{an}满足a1=1,a2=2,an=
1
3
(an-1+2an-2)(n∈N*且n≥3)
bn=
1n为奇数
-1n为偶数

(1)求an
(2)若cn=nanbn,n∈N*,求{cn}的前n项和Sn
题型:解答题难度:中档来源:不详

答案

(1)由an=class="stub"1
3
(an-1+2an-2)
得,an-an-1=-class="stub"2
3
(an-1-an-2)

又a2-a1=1≠0∴
an-an-1
an-1-an-2
=-class="stub"2
3
(n∈N*,n≥3)

∴数列{an+1-an}是首项为1,公比为-class="stub"2
3
的等比数列…(3分)∴an+1-an=(-class="stub"2
3
)n-1

从而,an-an-1=(-class="stub"2
3
)n-2
an-1-an-2=(-class="stub"2
3
)n-3
…a2-a1=1
以上各式相加得,an-a1=1+(-class="stub"2
3
)+…+(-class="stub"2
3
)n-2=
1-(-class="stub"2
3
)
n-1
1+class="stub"2
3

an=class="stub"8
5
-class="stub"3
5
(-class="stub"2
3
)n-1
…(6分)
(2)∵bn=
1n为奇数
-1n为偶数
,且cn=nanbn,n∈N*
cn=
class="stub"8
5
n-class="stub"3
5
(-class="stub"2
3
)n-1n,n为奇数
-class="stub"8
5
n+class="stub"3
5
(-class="stub"2
3
)n-1n,n为偶数
…(8分)
又Sn=c1+c2+…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
+…+n×class="stub"8
5
)
-class="stub"3
5
[1×(class="stub"2
3
)
0
+2×(class="stub"2
3
)
1
+3×(class="stub"2
3
)
2
+…+n×(class="stub"2
3
)
n-1
]

=class="stub"4
5
(1+n)-class="stub"3
5
[1×(class="stub"2
3
)
0
+2×(class="stub"2
3
)
1
+3×(class="stub"2
3
)
2
+…+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
+…-n×class="stub"8
5
)
-class="stub"3
5
[1×(class="stub"2
3
)
0
+2×(class="stub"2
3
)
1
+3×(class="stub"2
3
)
2
+…+n×(class="stub"2
3
)
n-1
]

=-class="stub"4
5
n-class="stub"3
5
[1×(class="stub"2
3
)
0
+2×(class="stub"2
3
)
1
+3×(class="stub"2
3
)
2
+…+n×(class="stub"2
3
)
n-1
]
…(10分)
令Tn=1×(class="stub"2
3
)0+2×(class="stub"2
3
)1+3×(class="stub"2
3
)2+…+n×(class="stub"2
3
)n-1
…(1)
class="stub"2
3
Tn
=1×(class="stub"2
3
)1+2×(class="stub"2
3
)2+3×(class="stub"2
3
)3+…+n×(class="stub"2
3
)n
…(2)
则由(1)(2)得,class="stub"1
3
Tn
=1+(class="stub"2
3
)+(class="stub"2
3
)2+(class="stub"2
3
)3+…+(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

Tn=9-(9+3n)(class="stub"2
3
)n

Sn=
class="stub"4n-23
5
+class="stub"27+9n
5
(class="stub"2
3
)n,n为奇数
-class="stub"4n+27
5
+class="stub"27+9n
5
(class="stub"2
3
)n,n为偶数
…(16分)

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