已知数列{An}的前n项和为Sn,A1=A2=1,bn=nSn+（n+2)An,数列{bn}是公差为d的等差数列,

易知b1=4,b2＝8,因此bn＝4n,得4＝sn+(n+2)/n*a(n)=sn+(n+2)/n*(sn-s(n-1)),因此sn＝(n+2)/(2n+2)*s(n-1)+2n/(n+1),易用归纳法证明sn0知an>0.注意到s4=13/4>3,于是sn>3,当n>=4时,故an