数列{an}中,a1=-1,a2=0,a(n+1)+4a(n-1)=4an(n>=2),数列bn满足 bn=a（n+1）-2an （1）证数列bn为等比数列,

a(n+1)+4a(n-1)=4an
a(n+1)-2an=2an-4a(n-1)
a(n+1)-2an=2[an-2a(n-1)]
bn=2b(n-1)
bn/b(n-1)=2
所以bn 是以2为公比的等比数列
b1=a2-2a1
=0-2*(-1)
=2
bn=b1q^(n-1)
=2*2^(n-1)
=2^n
a(n+1)-2an =2^n[两边同除以 2^(n+1)]
a(n+1)/2^(n+1)-2an/2^(n+1) =2^n/2^(n+1)
a(n+1)/2^(n+1)-2an/2^(n+1) =1/2
a(n+1)/2^(n+1)-2an/(2*2^n) =1/2
a(n+1)/2^(n+1)-an/2^n =1/2
所以an/2^n是以1/2为公差的等差数列
an/2^n=a1/2^1+(n-1)d
an/2^n=-1/2+(n-1)/2
an/2^n=n/2-1
an=n*2^(n-1)-2^n
an=(n-2)*2^(n-1)
sn=(-1)*2^0+0*2^1+1*2^2.+(n-2)*2^(n-1)
2sn=-1*2^1+0*2^2+1*2^3+.+(n-3)*2^(n-1)+(n-2)*2^n
sn-2sn=(-1)*2^0+2^1+2^2+.+2^(n-1)-(n-2)*2^n
-sn=-1+2*[1-2^(n-1)]/(1-2)-(n-2)*2^n
-sn=-1-2*[1-2^(n-1)]-(n-2)*2^n
-sn=-1-2+2^n-(n-2)*2^n
-sn=2^n-(n-2)*2^n-3
sn=(n-2)*2^n-2^n+3
sn=(n-3)*2^n+3