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.
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.
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=