题目

已知各项都不相等的等差数列{an}的前6项和为60,且a6为a1和a21的等比中项. (1)求数列{an}的通项公式; (2)若数列{bn}满足bn+1－bn=an(n∈N*),且b1=3,求数列的前n项和Tn. 答案：解:(1)设等差数列{an}的公差为d(d≠0),则解得∴an=2n+3. (2)由bn+1-bn=an,∴bn-bn-1=an-1(n≥2,n∈N*), bn=(bn-bn-1)+(bn-1-bn-2)+…+(b2-b1)+b1=an-1+an-2+…+a1+b1=(n-1)·+3=n(n+2). ∴bn=n(n+2)(n∈N*). ∴, Tn=

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