1、a(n+1）=（n+2)Sn/n,n=1,2,3...
=>S(n+1)-Sn=(n+2)Sn/n,=1,2,3...
=>S(n+1)=2(n+1)Sn/n,n=1,2,3...
=>[S(n+1)/(n+1)]/[Sn/n]=2,n=1,2,3...
所以,{Sn/n}是首项为S1/1=a1/1=1,公比为2的等比数列.
2、因为Sn/n=1*2^(n-1)
=>Sn=n*2^(n-1)
=>Sn-S(n-1)=an=(n+1)*2^(n-2),n=2,3,4...
又因为a1=1,也满足上式,
所以,综上述,有an=(n+1)*2^(n-2),n=1,2,3..