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an=a1.q^(n-1)
S2=2
a1(1+q) = 2 (1)
S3=-6
a1(1+q+q^2)=-6 (2)
(2)/(1)
(1+q+q^2)/(1+q)=-3
1+q+q^2 = -3-3q
q^2+4q+4 =0
q=-2
from (1)
a1(1+q) = 2
a1(1-2)=2
a1=-2
ie
an = (-2)^n
(2)
Sn= -(2/3)( 1- (-2)^n )
S(n+1)+S(n+2)
=-(2/3)( 1- (-2)^(n+1) ) -(2/3)( 1- (-2)^(n+2) )
=-(2/3)[ (-2)^(n+1) + (-2)^(n+2) ]
=-(2/3)(-2)^n. ( -2 + 4 )
=-(2/3)(-2)^n. 2
=2Sn
=>
S(n+1), Sn, S(n+2) 成等差数列
S2=2
a1(1+q) = 2 (1)
S3=-6
a1(1+q+q^2)=-6 (2)
(2)/(1)
(1+q+q^2)/(1+q)=-3
1+q+q^2 = -3-3q
q^2+4q+4 =0
q=-2
from (1)
a1(1+q) = 2
a1(1-2)=2
a1=-2
ie
an = (-2)^n
(2)
Sn= -(2/3)( 1- (-2)^n )
S(n+1)+S(n+2)
=-(2/3)( 1- (-2)^(n+1) ) -(2/3)( 1- (-2)^(n+2) )
=-(2/3)[ (-2)^(n+1) + (-2)^(n+2) ]
=-(2/3)(-2)^n. ( -2 + 4 )
=-(2/3)(-2)^n. 2
=2Sn
=>
S(n+1), Sn, S(n+2) 成等差数列
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1）S2=a1(1+q)=2
a3=S3-S2=a1q^2=-8
综合两式得q=-2,a1=-2
2)Sn=-2/3(1-(-2)^n)
S(n+2)+S(n+1)-2Sn=1/3*(-2)^n(2-2)=0
故Sn+1、Sn、Sn+2是等差数列
a3=S3-S2=a1q^2=-8
综合两式得q=-2,a1=-2
2)Sn=-2/3(1-(-2)^n)
S(n+2)+S(n+1)-2Sn=1/3*(-2)^n(2-2)=0
故Sn+1、Sn、Sn+2是等差数列
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