设an=aq^(n-1)
sn=a(1-q^n)/(1-q)
1.
s4=a(1-q^4)/(1-q)
s10=a(1-q^10)/(1-q)
s7=a(1-q^7)/(1-q)
s4*s10=(s7)^2
[a(1-q^4)/(1-q)][a(1-q^10)/(1-q)]=[a(1-q^7)/(1-q)]^2
(1-q^4)(1-q^10)=(1-q^7)^2
1-q^4-q^10+q^14=1-2q^7+q^14
q^4+q^10=2q^7
1+q^6=2q^3
a+aq^6=2aq^3
a1+a7=2a4
所以a1,a7,a4也成等比数,证毕.
2. 设S3=3/2,S6=21/16,bn=X*an-n*n若{bn}数列是单调递减数列,求实数X的取值范
3/2=S3=a(1-q^3)/(1-q)
21/16=S6=a(1-q^6)/(1-q)
后式除以前式：
(1-q^6)/(1-q^3)=(21/16)/(3/2)
1+q^3=7/8
q^3=-1/8
q=-1/2
代入上式可得a=2
所以an=2(-1/2)^(n-1)=-(-1/2)^(n-2)
bn=X*an-n*n
=-x(-1/2)^(n-2)-n^2
b(n-1)=-x(-1/2)^(n-3)-(n-1)^2
bn-b(n-1)=[-x(-1/2)^(n-2)-n^2]-[-x(-1/2)^(n-3)-(n-1)^2]
=-x(-1/2)^(n-2)-n^2-2x(-1/2)^(n-2)+(n-1)^2
=-x3(-1/2)^(n-2)-2n+1
＜0
x3(-1/2)^(n-2)+2n-1＞0
当n为偶数时,不妨设n=2k
x3(-1/2)^(2k-2)+4k-1＞0
x3(1/2)^(2k-2)+4k-1＞0
x＞(1-4k)/[3(1/2)^(2k-2)]
=(1-4k)2^(2k-2)/3
＞(1-4*1)2^(2*1-2)/3
=-1
当n为奇数时,不妨设n=2m+1
x3(-1/2)^(2m-1)+4m+1＞0
-x3(1/2)^(2m-1)+4m+1＞0
x＜(1+4m)/[3(1/2)^(2m-1)]
=(1+4m)2^(2m-1)/3
＜(1+4*1)2^(2*1-1)/3
=10/3
所以-1＜x＜10/3