Question

已知數(shù)列{an}是等差數(shù)列,a2=6,a5=12,數(shù)列{bn}的前n項(xiàng)和是Sn,且Sn+bn=1.

(1)求數(shù)列{an}的通項(xiàng)公式.

(2)求證:數(shù)列{bn}是等比數(shù)列.

(3)記cn=,{cn}的前n項(xiàng)和為Tn,若Tn<對(duì)一切n∈N*都成立,求最小正整數(shù)m.

 

Answer 1

(1) an=2n+2 (2)見(jiàn)解析 (3) 2012

【解析】(1)設(shè){an}的公差為d,則a2=a1+d,a5=a1+4d.

∵a2=6,a5=12,∴

解得:a1=4,d=2.∴an=4+2(n-1)=2n+2.

(2)當(dāng)n=1時(shí),b1=S1,由S1+b1=1,得b1=.

當(dāng)n≥2時(shí),∵Sn=1-bn,Sn-1=1-bn-1,

∴Sn-Sn-1=(bn-1-bn),即bn=(bn-1-bn).

∴bn=bn-1.

∴{bn}是以為首項(xiàng),為公比的等比數(shù)列.

(3)由(2)可知:bn=·()n-1=2·()n.

∴cn====-,

∴Tn=(1-)+(-)+(-)+…+(-)=1-<1,

由已知得≥1,∴m≥2012,

∴最小正整數(shù)m=2012.