Question

1.设Sn是数列an的前n项和,a1=a,且Sn^2=3*n^2*an+（Sn-1)^2,an不等于0,n=2,3,4.

Answer 1

1.Sn^2=3*n^2*an+（Sn-1)^2

[Sn+Sn-1][Sn-Sn-1]=3n^2*an

[Sn+Sn-an]*an=3n^2*an

Sn+Sn -an=3n^2

Sn=3n^2/2+an/2

S(n+1)=3(n+1)^2/2+a(n+1)/2

a(n+1)=3(2n+1)/2+a(n+1)/2-an/2

a(n+1)=6n+3-an

a(n+2)=6n+9-a(n+1)

=6n+9-6n-3+an=6+an

所以a(n+2)-an=6

2.4Sn=an^2+4n-1

4S(n-1)=a(n-1)^2+4n-5

4an=an^2-a(n-1)^2+4

an^2-4an+4-a(n-1)^2=0

(an - 2)^2-a(n-1)^2=0

4a1=a1^2+3

a1=1或3

an -2 =a(n-1)或an-2=-a(n-1)

所以an=2n-1或2n+1或1