天津市高考數(shù)學(xué)文二輪復(fù)習(xí)題型練大題專項(xiàng)數(shù)列的通項(xiàng)求和問(wèn)題

1、題型練題型練4大題專項(xiàng)大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問(wèn)題數(shù)列的通項(xiàng)、求和問(wèn)題1.設(shè)數(shù)列an的前n項(xiàng)和為Sn滿足(1q)Snqan=1且q(q1)≠0.(1)求an的通項(xiàng)公式(2)若S3S9S6成等差數(shù)列求證:a2a8a5成等差數(shù)列.2.已知等差數(shù)列an的首項(xiàng)a1=1公差d=1前n項(xiàng)和為Snbn=.(1)求數(shù)列bn的通項(xiàng)公式(2)設(shè)數(shù)列bn前n項(xiàng)和為Tn求Tn.(2)求數(shù)列|bn|的前n項(xiàng)和Tn.6.記U=12…100.對(duì)數(shù)列an(n∈N

2、)和U的子集T若T=?定義ST=0若T=t1t2…tk定義ST=….例如:T=1366時(shí)ST=a1a3a66.現(xiàn)設(shè)an(n∈N)是公比為3的等比數(shù)列且當(dāng)T=24時(shí)ST=30.(1)求數(shù)列an的通項(xiàng)公式(2)對(duì)任意正整數(shù)k(1≤k≤100)若T?12…k求證:STak1(3)設(shè)C?UD?USC≥SD求證:SCSC∩D≥2SD.##題型練4大題專項(xiàng)(二)數(shù)列的通項(xiàng)、求和問(wèn)題1.解(1)當(dāng)n=1時(shí)由(1q)S1qa1=1a1=1.當(dāng)n≥2時(shí)由

3、(1q)Snqan=1得(1q)Sn1qan1=1兩式相減得an=qan1.又q(q1)≠0所以an是以1為首項(xiàng)q為公比的等比數(shù)列故an=qn1.(2)由(1)可知Sn=又S3S6=2S9所以化簡(jiǎn)得a3a6=2a9兩邊同除以q得a2a5=2a8.故a2a8a5成等差數(shù)列.2.解(1)∵在等差數(shù)列an中a1=1公差d=1∴Sn=na1d=∴bn=.(2)bn==2∴Tn=b1b2b3…bn=2…=2…=2.故Tn=.3.證明(1)因?yàn)閍n