Question

已知数列{a n }是等差数列,a 2 =6,a 5 =12,数列{b n }的前n项和是S n ,且S n + b n

Answer 1

(1) a _n=2n+2 (2)见解析 (3) 2012

(1)设{a _n}的公差为d,则a ₂=a ₁+d,a ₅=a ₁+4d.

∵a ₂=6,a ₅=12,∴

解得:a ₁=4,d=2.∴a _n=4+2(n-1)=2n+2.

(2)当n=1时,b ₁=S ₁,由S ₁+

b ₁=1,得b ₁=

.

当n≥2时,∵S _n=1-

b _n,S _n-1=1-

b _n-1,

∴S _n-S _n-1=

(b _n-1-b _n),即b _n=

(b _n-1-b _n).

∴b _n=

b _n-1.

∴{b _n}是以

为首项,

为公比的等比数列.

(3)由(2)可知:b _n=

·(

) ^n-1=2·(

) ⁿ.

∴c _n=

=

=

=

-

,

∴T _n=(1-

)+(

-

)+(

-

)+…+(

-

)=1-

<1,

由已知得

≥1,∴m≥2012,

∴最小正整数m=2012.