由,An+A(n+1)=4n可得,n≥2时,A(n-1)+An=4(n-1)两式相减,得A(n+1)-A(n-1)=4所以,n≥2时数列{An}的奇数项和偶数项分别为公差为4的等差数列由,A1＋A2＝4,A1＝1可得,A2＝3设n为偶数,则n-1为奇数An＝3＋[(n-2)/2+1-1]×4...