n>=3,
b(n-1)=a1^2+a2^2+…+a(n+1)^2-a1a2…a(n+1)
b(n-1)-b(n-2)=a(n+1)^2-a1a2...a(n+1)+a1..an
=a(n+1)[a(n+1)-a1...an]+a1...an
=-a(n+1)+a1..an
=1
因此bn为公差为1的等差数列
b1=a1^2+a2^2+a3^2-a1a2a3=12-8=4
所以bn=3+n
.