a(n+1)=[a(n)-1]/[a(n)+3],
a(n+1)+1=[a(n)-1]/[a(n)+3] +1=[2a(n)+2]/[a(n)+3]=2[a(n)+1]/[a(n)+3],
若a(n+1)+1=0,则a(n)+1=0,...,a(1)+1=0,与a(1)=-1/3矛盾.
因此,a(n)+1 不为0.
1/[a(n+1)+1] = [a(n)+3]/[2a(n)+2] = [a(n)+1+2]/[2a(n)+2] = 2/[2a(n)+2] + [a(n)+1]/[2a(n)+2]
=1/[a(n)+1] + 1/2,
{1/[a(n)+1]}是首项为1/[a(1)+1]=1/[1-1/3]=3/2,公差为1/2的等差数列.
1/[a(n)+1] = 3/2 +(n-1)/2 = (n+2)/2,
a(n)+1=2/(n+2),
a(n)=2/(n+2) - 1 = -n/(n+2)