s(n)=(n+1)[a(n)+1]/2-1.
s(n+1)=(n+2)[a(n+1)+1]/2-1,
a(n+1)=s(n+1)-s(n)=[(n+2)a(n+1)-(n+1)a(n)]/2,
na(n+1)=(n+1)a(n),
a(n+1)/(n+1)=a(n)/n,
{a(n)/n}为首项为a(1)/1=3,的常数数列.
a(n)/n=3,
a(n)=3n=3+3(n-1),
{a(n)}是首项为3,公差为3的等差数列.
楼主英明。。。a(n+1)=s(n+1)-s(n)=[(n+2)a(n+1)-(n+1)a(n)+1]/2,na(n+1)=(n+1)a(n)+1,a(n+1)/(n+1)=a(n)/n+1/[n(n+1)]=a(n)/n+1/n-1/(n+1),a(n+1)/(n+1)+1/(n+1)=a(n)/n+1/n.{a(n)/n+1/n}为首项为a(1)/1+1=4,的常数数列。a(n)/n+1/n=4,a(n)=4n-1=4(n-1)+3,{a(n)}是首项为3,公差为4的等差数列。