Sn=4/3An-1/3x2的n-1次方+2/3,n=1,2 … 已知{An+2的n次方}为等比数列,求An=?

Sn=(4/3)*An-(1/3)*2^(n-1)+2/3
S(n+1)=(4/3)*A(n+1)-(1/3)*2^n+2/3
S1=A1=(4/3)*A1-(1/3)+2/3
A1=-1
A(n+1)=S(n+1)-Sn=(4/3)*[A(n+1)-An)-(1/6)*2^n
A(n+1)=(4/3)*an+(1/6)*2^n
A(n+1)-(1/4)*2^(n+1)=(4/3)*[An-(1/4)*2^n]
∴An-(1/4)*2^n是比为4/3的等比数列
An-(1/4)*2^n=[A1-(1/4)*2]*(4/3)^(n-1)
An=2^(n-2)-2*(4/3)^(n-2) .(1)
An+2^n=5*2^(n-2)-2*(4/3)^(n-2)
因此An+2^n不可能是等比数列.

a(n+1)+n+1+2=2(an+n+2) 令bn=an+n+2 有，b1=4 b(n+1)=2bn 所以，bn=4*2^(n-1)=2^(n+1) 所以，an=2^(n+1)-n-2 2）要