1.设{an}是公比为q的等比数列,|q|>1,令bn=an+1(n=1,2,…),若数列{bn}有连接四项在集合{-53,-23,19,37,82}中,则6q=?2.数列｛an｝的通项an=n^2(cos^2(nπ／3)-sin^2(nπ／3)),其前n项和为Sn,则S30为?

Question

1.设{an}是公比为q的等比数列,|q|>1,令bn=an+1(n=1,2,…),若数列{bn}有连接四项在集合{-53,-23,19,37,82}中,则6q=?
2.数列｛an｝的通项an=n^2(cos^2(nπ／3)-sin^2(nπ／3)),其前n项和为Sn,则S30为?

Answer 1

1.若数列{bn}有连续四项在集合{-53,-23,19,37,82}中
则：若数列{an}有连续四项在集合{-54,-24,18,36,81}
则这4个数必是18 -24 36 -54 公比为-9
6q=-54
2.(cos(nπ/3))^2-(sin(nπ/3))^2=cos(2nπ/3)
n=1,cos(2π/3)=-1/2
n=2,cos(4π/3)=-1/2
n=3,cos(6π/3)=1
以后cos取值三个一组循环.
三个一组分析,3n-2、3n-1、3n为一组
(3n)^2×(1)=(3n)^2×(1/2)+(3n)^2×(1/2)
分别与(3n-2)^2×(-1/2)和(3n-1)^2×(-1/2)相加
(3n)^2×(1/2)+(3n-2)^2×(-1/2)
=(1/2)×((3n)^2-(3n-2)^2)
=(1/2)×(3n+(3n-2))×(3n-(3n-2))
=(1/2)×(6n-2)×(2)
=6n-2
另一个同理
(3n)^2×(1/2)+(3n-1)^2×(-1/2)
=(1/2)×((3n)^2-(3n-1)^2)
=(1/2)×(3n+(3n-1))×(3n-(3n-1))
=(1/2)×(6n-1)×(1)
=3n-0.5
(3n-2)^2×(-1/2)+(3n-1)^2×(-1/2)+(3n)^2×(1)
=(1/2)×[(3n)^2-(3n-2)^2]+(1/2)×[(3n)^2-(3n-1)^2]
=(1/2)(3n+3n-2)(3n-3n+2)+(1/2)(3n+3n-1)(3n-3n+1)
=9n-2.5
也可以直接展开
Bn=9n-2.5
B1=A1+A2+A3 B2=A4+A5+A6 ……
S30共10组
S30=B1+B2+……+B10
用等差数列求和公式
B1=9×1-2.5=6.5
B10=9×10-2.5=87.5
S30=(B1+B10)×10/2=(6.5+87.5)×10/2=470
有不明白的百度I我,不常在