a(n)*a(n+1)=a(n)-3a(n+1)+8 证明{a(n)+4/a(n)-2}是等比数列,并求{an}的通项公

a(n)*a(n+1)=a(n)-3a(n+1)+8
a(n+1)(an+3) = an +8
a(n+1) = (an+8)/(an+3)
The aux. equation
x^2+2x-8=0
x= -4 or 2
a(n+1) = (an+8)/(an+3)
a(n+1)-2 = (an+8)/(an+3)-2
= (-an+2)/(an+3)
1/[a(n+1) -2] = (an+3)/(-an+2)
= - (1+ 5/(an-2))
1/[a(n+1) -2]+1/6 = -5 ( 1/(an-2) + 1/6)
[1/[a(n+1) -2]+1/6]/( 1/(an-2) + 1/6) = -5
( 1/(an-2) + 1/6)/( 1/(a1-2) + 1/6) = (-5)^(n-1)
1/(an-2) + 1/6= (-5/6)(-5)^(n-1)
1/(an-2) = -1/6 +(1/6)(-5)^n
an -2 = 6/[-1+(-5)^n]
an = 2+ 6/[-1+(-5)^n]

a(n)*a(n+1)=a(n)-3a(n+1)+8
a(n+1)*(a(n)+3)=a(n)+8
a(n+1)=(a(n)+8)/(a(n)+3)
(a(n+1)+4)/(a(n+1)-2)
=[((a(n)+8)/(a(n)+3))+4]/[((a(n)+8)/(a(n)+3))-2]
=[(a(n)+8)+4*(a(n)+3))]/[(a(n)+8)-2*(a(n)+3))]
=-5*(a(n)+4)/(a(n)-2)
所以{a(n)+4/a(n)-2}成等比数列，公比为-5
。。。。。