因为a(n+1)=an+n2+2n,an=a(n-1)+(n-1)2+2(n-1)...a2=a1+1+2,累加得a(n+1)+an+...+a2=an+a(n-1)+...+a1+n2+2n+(n-1)2+2(n-1)+...+1+2,即an-a1=[n(n+1)(n+2)]\6+n2+2n,所以an=n3+4n2+2n+22