Sn^2=3+(n-1)=n+2
所以 Sn=√(n+2)
a1=S1=√3
所以有 an=Sn-S[n-1]=√(n+2)-√(n+1) n>=2
n>=2时 我们来比较
Sn与3na[n]
令f(n)=Sn-3na[n]
=√(n+2)-3ncc(n+2)+3n√(n+1)
=3n√(n+1)-(3n-1)√(n+2)
=√(9n^3+9n^2)-√(9n^3+12n^2-11n+2)
只需要比较:
(9n^3+9n^2)-(9n^3+12n^2-11n+2)
=11n-3n^2-2
容易判断出 2=4时 f(n)