问题
解答题
数列{an}的通项an=n2(cos2
(1)求S3、S6的值 (2)求前3n项的和S3n (3)若bn=
|
答案
(1)an=n2(cos2
-sin2nπ 3
)nπ 3
=n2cos
,n∈N*,2nπ 3
cos
以3为周期.2nπ 3
∴S3=a1+a2+a3
=cos
+22cos2π 3
+32cos4π 3 6π 3
=-
+4×(-1 2
) +9×1=1 2
.13 2
S6=(a1+a2+a3)+(a4+a5+a6)
=[-
+4×(-1 2
)+9×1]+[16×(-1 2
)+25×(- 1 2
) +36×1]1 2
=22.
(2)∵a3n-2+a3n-1+a3n
=(3n-2)2•(-
) +(3n-1)2•(-1 2
) +(3n)2•1=9n-1 2
,5 2
∴S3n=(a1+a2+a3)+(a4+a5+a6)+…+(a3n-2+a3n-1+a3n)
=(9-
)+(9×2-5 2
)+…+(9n-5 2
)5 2
=9(1+2+…+n)-
=5n 2
.(9分)9n2+4n 2
(3)bn=
=S3n n•4n
•(9n+4 2
)n,1 4
∴Tn=
(1 2
+13 4
+22 42
+…+31 43
),9n+4 4n
∴4Tn=
(13+1 2
+22 4
+…+31 42
),9n+4 4n-1
∴3Tn=
(13+1 2
+9 4
+…+9 42
-9 4n-1
)9n+4 4n
=8-
-1 2 2n-3
,9n 22n+1
∴Tn=
-8 3
-1 3•22n-3
.(14分).3n 2 2n+1