等比数列An,a1+a2+a3=-3,a1a2a3=8求此数列前n项和Sn

a1a2a3=8
(a2)^3=8
a2=2
a1+a2+a3=-3
a1+2+a3=-3
a1+a3=-5
a2/q+a2q=-5
2/q+2q+5=0
2q^2+5q+2=0
(2q+1)(q+2)=0
q=-1/2或q=-2
当q=-1/2时
a2=a1q
2=a1*(-1/2)
a1=-4
an=a1q^(n-1)
an=-4*(-1/2)^(n-1)
an=-(-1/2)^(-2)*(-1/2)^(n-1)
an=-(-1/2)^(n-3)
Sn=a1(1-q^n)/(1-q)
=-4*[1-(-1/2)^n]/(1+1/2)
=-4*2/3*[1-(-1/2)^n]
=-8/3+8/3*(-1/2)^n
当q=-2时
a2=a1q
2=a1*(-2)
a1=-1
an=a1q^(n-1)
an=-(-1/2)^(n-1)
Sn=a1(1-q^n)/(1-q)
=-[1-(-2)^n]/(1+2)
=-1/3*[1-(-2)^n]
=-1/3+1/3*(-2)^n

设a2=n,a1=n/q,a3=nq
a1a2a3=8:n³=8,n=2
∴a2=2
∴a1a3=4,
∵等比数列,a1+a2+a3=-3＜0
∴可以看出a1=-1,a3=-4,q=-2
Sn=a1(1-q^n)/(1-q)
=-1(1+2^n)/3
=-1/3(1+2^n)