1.在等比数列{a}中,前几项和为Sn(1)a7=12,q=根号2开3次方,求a19(2)a2=-2,a5=54,求a8(3)a1=5,a9a10=100,求a18(4)S3+S6=2S9,求数列的公比q.

问题描述:

1.在等比数列{a}中,前几项和为Sn(1)a7=12,q=根号2开3次方,求a19(2)a2=-2,a5=54,求a8(3)a1=5,a9a10=100,求a18(4)S3+S6=2S9,求数列的公比q.
2.在等比数列{an}(n属于N+)中,若a1=1,a4=1/8,求该数列的前10项和.
3.已知正项等比数列{an}中,a1=8设bn=log2(an)(n属于N+).
(1)求证:数列{bn}是等差数列;(2)如果数列{bn}的前7项和S7是它的前n项和Sn的最大值,且S7不等于S8,S7不等于S6,求数列{an}的公比q的取值范围.

1、解1)∵a7=a1*q^6=12,
a19=a1*q^18=a1*q^6*q^12
=12*[√2^(1/3)]^18=12*2^3=12*8=96
2)a2=a1q=-2,a5=a1*q^4=54
q^3=-27,∴q=-3,a1=2/3
∴a8=a1*q^7=2/3*(-3)^7=-1458
3)a9*a10=a1*a18
∴a18=a9*a10/a1=100/5=20
4)∵S3+S6=2S9
∴ a1(1-q^2)/(1-q)+a1*(1-q^5)/(1-q)
=2a1(1-q^8)/(1-q)
即1-q^2+1-q^5=2-2*q^8
q^2*(2q^6-q^3-1)=0
q^3=1,q^3=-1/2
q=0,q=1,舍去
∴q=(-1/2)^(1/3)
2、∵a4=a1*q^3=1/8,a1=1,
∴q^3=1/8,q=1/2,
∴S10=a1*(1-q^9)/(1-q)
=1*[1-(1/2)^9]/(1-1/2)
=511/256
1)∵bn=log2(an),得b1=log2(8)=3
∴bn-b(n-1)=log2(an)-log2[a(n-1)]
=log2[an/a(n-1)]=log2(q)=定值
则{bn}是首项a1=3,公差d=log2(q)的等差数列
2)、由上可知
Sn=nb1+[n(n-1)log2(q)]/2
=3n+[n(n-1)log2(q)]/2
S6=18+15log2(q),S7=21+21log2(q)
S8=24+28log2(q)
∵S7>S6,S7>S8,
∴21+21*log2(q)>18+15log2(q),
21+21*log2(q)>24+28log2(q)
即-1/2<log2(q)<-3/7
∴√2/2<q<2^(-3/7)