Question

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.

 

Answer 1

(1) an=2n+ (2) Tn=

【解析】(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(n+2),

当n=1时,b1=3也适合上式,

∴bn=n(n+2)(n∈N*).

∴==(-),

Tn=(1-+-+…+-)

=(--)=.