Question

已知各項(xiàng)都不相等的等差數(shù)列{an}的前6項(xiàng)和為60,且a6為a1和a21的等比中項(xiàng).

(1)求數(shù)列{an}的通項(xiàng)公式.

(2)若數(shù)列{bn}滿(mǎn)足bn+1-bn=an(n∈N*),且b1=3,求數(shù)列{}的前n項(xiàng)和Tn.

 

Answer 1

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

【解析】(1)設(shè)等差數(shù)列{an}的公差為d(d≠0),

則

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),

當(dāng)n=1時(shí),b1=3也適合上式,

∴bn=n(n+2)(n∈N*).

∴==(-),

Tn=(1-+-+…+-)

=(--)=.