关于等差数列等比数列的计算问题.1.已知数列an满足an+1=an+3n+2,且a1=2,求an0 因为an+1=an+3n+2,所以an-an-1=3n-1(n>=2)所以an=(an-an-1)+(an-1-an-2)+……+(a2-a1)+a1=3n-1+3n-4+……+5+2请问,an-an-1不是应该等于d吗?
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