已知数列{an}的前n项和Sn=n2-6n,令bn=ancos2nπ3,记数列{bn}的前项和为Tn,则T31=______.-数学

题目简介

已知数列{an}的前n项和Sn=n2-6n,令bn=ancos2nπ3,记数列{bn}的前项和为Tn,则T31=______.-数学

题目详情

已知数列{an}的前n项和Sn=n2-6n,令bn=ancos
2nπ
3
,记数列{bn}的前项和为Tn,则T31=______.
题型:填空题难度:中档来源:不详

答案

∵数列{an}的前n项和Sn=n2-6n,
∴当n≥2时,an=Sn-Sn-1=2n-7;
当n=1时,a1=S1=-5,也符合上式;
∴an=2n-7;
又bn=ancosclass="stub"2nπ
3

∴当n=1时,b1=-class="stub"1
2
a1=-class="stub"1
2
×(-5)=class="stub"5
2

同理可得,b2=-class="stub"1
2
a2=class="stub"3
2

b3=a3=-1,
∴b1+b2+b3=3;
同理可得,b4+b5+b6=3,
b7+b8+b9=3,

又b31=-class="stub"1
2
a31=-class="stub"1
2
×(2×31-7)=-class="stub"55
2

∴数列{bn}的前31项和为T31=(b1+b2+b3)+(b4+b5+b6)+…+(b28+b29+b30)+b31
=3×10+b31
=30-class="stub"55
2

=class="stub"5
2

故答案为:class="stub"5
2

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