解:
由于: an-a(n-1)=a(n-1)/n+(n+1)
则：an=[(n+1)/n]*a(n-1)+(n+1)
则：an/(n+1)=a(n-1)/n +1
设bn=an/(n+1)
则：b(n) -b(n-1)=1
则{bn}为公差为1的等差数列
则：bn=b1+(n-1)*1=a1/2 +(n-1)=n+2
由于：bn=an/(n+1)=(n+2)
则：an=(n+1)(n+2)=n^2+3n+2
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