设E项数列{an}满足A1=1，An+1等于8An的四次方根，求An

解: 由于a(n+1)=[8an]^(1/4)
则:两边同时取以 8 为底的对数,
得:
log8[a(n+1)]=log8[8an^(1/4)]
log8[a(n+1)]=(1/4)*log8[8an]
log8[a(n+1)]=(1/4)*[log8(8)+log8(an)]
log8[a(n+1)]=(1/4)*[1+log8(an)]
设bn=log8[an]
则: b(n+1)=(1/4)+(1/4)bn
b(n+1)-1/3=(1/4)[bn-1/3]
[b(n+1)-1/3]/[bn-1/3]=1/4
则:bn-1/3为公比为 1/4 的等比数列
则: bn-1/3=(b1-1/3)*(1/4)^(n-1)
=(-1/3)*(1/4)^(n-1)
=(-4/3)*(1/4)^n
则:bn=(-4/3)*(1/4)^n+1/3
则:bn=log8[an]=(-4/3)*(1/4)^n+1/3
则: an=8^[(-4/3)*(1/4)^n+1/3]
=2*8^[(-4/3)*(1/4)^n]

An=2^(((1/2)^(2n-2))-1)

由A(n+1)=8(An)^(1/4)得
2A(n+1)=16(An)^(1/4)=(2(An))^(1/4),即2A(n+1)=(2An)^(1/4),
故2A2=2^(1/4),2A3=(2A2)^(1/4),2A4=(2A3)^(1/4)
....
于是得
2An=(2A(n-1)))^(1/4)=(2A(n-2)))^((1/4)^2)=(2A(n-3)))^((1/4)^3)=...
=(2A1)^((1/4)^(n-1))=2^((1/4)^(n-1))=2^((1/2)^(2n-2))
即
An=2^((1/2)^(2n-2))/2=2^((