第二问
由已知得,当n=1时,a13=S12=a12,
又∵an>0,∴a1=1
当n≥2时,a13+a23++an3=Sn2①
a13+a23++an-13=Sn-12②
由①-②得,an3=Sn2-Sn-12=(Sn-Sn-1)(Sn+Sn-1)=an(Sn+Sn-1)
∴an2=Sn+Sn-1=2Sn-an(n≥2)
显然当n=1时,a1=1适合上式.
故an2=2Sn-an(n∈N*)③
an-12=2Sn-1-an-1(n≥2)④
由③-④得,an2-an-12=2Sn-2Sn-1-an+an-1=an+an-1
∵an+an-1>0∴an-an-1=1(n≥2)
故数列an是首项为1,公差为1的等差数列.
∴an=n(n∈N*)