a(n+1)=an^2+2an
a(n+1)+1=(an+1)^2
log2[(a(n+1)+1]=2log2[(an)+1]
log2[(a(n+1)+1]/log2[an+1]=2
{log2[a(n+1)+1]}等比数列,公比：2,首项：1
log2[(an)+1]=2^(n-1)
an=2^[2^(n-1)]-1=4^(n-1)-1
a(n+1)=an^2+2an
bn=[(an)+1]/(an+1),(分子分母同乘an)
=>bn=[(an+1)-(an)]/(an)(an+1)=1/an-1/a(n+1)