数列{an}的通项an=n2(cos2nπ3-sin2nπ3),其前n项和为Sn,(1)求Sn;(2)bn=S3nn•4n,求数列{bn}的前n项和Tn.-数学

题目简介

数列{an}的通项an=n2(cos2nπ3-sin2nπ3),其前n项和为Sn,(1)求Sn;(2)bn=S3nn•4n,求数列{bn}的前n项和Tn.-数学

题目详情

数列{an}的通项an=n2(cos2
3
-sin2
3
)
,其前n项和为Sn
(1)求Sn
(2)bn=
S3n
n•4n
,求数列{bn}的前n项和Tn
题型:解答题难度:中档来源:江西

答案

(1)由于cos2class="stub"nπ
3
-sin2class="stub"nπ
3
=cosclass="stub"2nπ
3
an=n2•cosclass="stub"2nπ
3

故S3k=(a1+a2+a3)+(a4+a5+a6)+…+(a3k-2+a3k-1+a3k)
=(-
12+22
2
+32)+(-
42+52
2
+62)+…+[-
(3k-2)2+(3k-1)2
2
+(3k)2]

=class="stub"13
2
+class="stub"31
2
+…+class="stub"18k-5
2
=
k(4+9k)
2

S3k-1=S3k-a3k=
k(4-9k)
2

S3k-2=S3k-1-a3k-1=
k(4-9k)
2
+
(3k-1)2
2
=class="stub"1
2
-k=-class="stub"3k-2
3
-class="stub"1
6

Sn=
-class="stub"n
3
-class="stub"1
6
n=3k-2
(n+1)(1-3n)
6
n=3k-1
n(3n+4)
6
n=3k
(k∈N*)
(2)bn=
S3n
n•4n
=class="stub"9n+4
2•4n

Tn=class="stub"1
2
[class="stub"13
4
+class="stub"22
42
++class="stub"9n+4
4n
]

4Tn=class="stub"1
2
[13+class="stub"22
4
++class="stub"9n+4
4n-1
]

两式相减得3Tn=class="stub"1
2
[13+class="stub"9
4
+…+class="stub"9
4n-1
-class="stub"9n+4
4n
]=class="stub"1
2
[13+
class="stub"9
4
-class="stub"9
4n
1-class="stub"1
4
-class="stub"9n+4
4n
]=8-class="stub"1
22n-3
-class="stub"9n
22n+1

Tn=class="stub"8
3
-class="stub"1
3•22n-3
-class="stub"3n
22n+1

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