《導(dǎo)學(xué)案》2021版高中數(shù)學(xué)(人教A版-必修5)教師用書(shū)：2.9等差、等比數(shù)列的綜合應(yīng)用-練習(xí)

1.數(shù)列{an}是首項(xiàng)為a1=1,公差為d=3的等差數(shù)列,假如an=2005,則序號(hào)n等于().A.667B.668C.669D.670【解析】由題意有an=1+3(n-1),假如an=2005,則1+3(n-1)=2005,故n=669.【答案】C2.已知由正數(shù)組成的等比數(shù)列{an}中,公比q=2,a1·a2·a3·…·a30=245,則a1·a4·a7·…·a28等于().A.25 B.210 C.215【解析】由等比數(shù)列的性質(zhì)有a2·a5·a8·…·a29=a1·a4·a7·…·a28·210,a3·a6·a9·…·a30=a1·a4·a7·…·a28·220,故a1·a4·a7·…·a28==25.【答案】A3.設(shè)數(shù)列{an},{bn}都是等差數(shù)列.若a1+b1=7,a3+b3=21,則a5+b5=.

【解析】由等差數(shù)列的性質(zhì)有a1+a5=2a3,b1+b5=2b3∴(a1+b1)+(a5+b5)=2(a3+b3),∴a5+b5=42-7=35.【答案】354.有兩個(gè)等差數(shù)列2,6,10,…,190及2,8,14,…,200,由這兩個(gè)等差數(shù)列的公共項(xiàng)按從小到大的挨次組成一個(gè)新數(shù)列,求這個(gè)新數(shù)列{an}的通項(xiàng)公式an.【解析】分析可得新數(shù)列仍為等差數(shù)列,且首項(xiàng)為2,又兩數(shù)列的公差分別為4和6,而新數(shù)列的公差為4和6的最小公倍數(shù)12,故an=2+(n-1)×12=12n-10.5.等比數(shù)列{an}滿足a1=1,且,,成等差數(shù)列,則數(shù)列{an}的前10項(xiàng)和為().A.10 B.20 C.256 D.【解析】設(shè)等比數(shù)列{an}的公比為q,由條件可得=+,即=+1,解之得q=1,故an=1,前10項(xiàng)的和為10.【答案】A6.已知數(shù)列{an}是首項(xiàng)為2,公差為1的等差數(shù)列,數(shù)列{bn}是首項(xiàng)為1,公比為2的等比數(shù)列,則數(shù)列{}前10項(xiàng)的和等于().A.511 B.512 C.1023 D.【解析】an=2+(n-1)×1=n+1,bn=1×2n-1=2n-1,依題意可令Mn=a1+a2+a4+…+=(1+1)+(2+1)+…+(2n-1+1)=2n-1+n,∴M10=210+10-1=1033.【答案】D7.若兩個(gè)等差數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,且滿足=,則=.

【解析】=====.【答案】8.在數(shù)列{an}中,a1=1,an+1=2an+2n.(1)設(shè)bn=,證明:數(shù)列{bn}是等差數(shù)列;(2)求數(shù)列{an}的前n項(xiàng)和Sn.【解析】(1)由an+1=2an+2n,兩邊同除以2n,得=+1.∴-=1,即bn+1-bn=1,∴{bn}是首項(xiàng)為1,公差為1的等差數(shù)列.(2)由(1)得,bn=n.∴an=n·2n-1,∴Sn=20+2×21+3×22+…+n×2n-1,①∴2Sn=21+2×22+…+(n-1)2n-1+n·2n.②由①-②得-Sn=20+21+22+…+2n-1-n·2n=-n·2n=(1-n)·2n-1,∴Sn=(n-1)·2n+1.9.同學(xué)們都有這樣的解題閱歷:在某些數(shù)列的求和中,可把其中一項(xiàng)分裂成兩項(xiàng)之差,使得某些項(xiàng)可以相互抵消,從而實(shí)現(xiàn)化簡(jiǎn)求和.如:已知數(shù)列{an}的通項(xiàng)為an=,則將其通項(xiàng)化為an=-,故數(shù)列{an}的前n項(xiàng)和Sn=(1-)+(-)+…+(-)=1-=.“斐波那契數(shù)列”是數(shù)學(xué)史上一個(gè)出名的數(shù)列,在斐波那契數(shù)列{an}中,a1=1,a2=1,an+an+1=an+2(n∈N*),若a2021=a,那么數(shù)列{an}的前2011項(xiàng)的和是.

【解析】由已知有a1+a2=a3,a2+a3=a4,…,a2011+a2012=a2013,各式相加,得a2+a1+a2+a3+…+a2011=a2013,故數(shù)列{an}的前2011項(xiàng)和為a-1.【答案】a-110.已知數(shù)列{an}為等差數(shù)列,且a1=1,{bn}為等比數(shù)列,數(shù)列{an+bn}的前三項(xiàng)依次為3,7,13.求(1)數(shù)列{an},{bn}的通項(xiàng)公式;(2)數(shù)列{anbn}的前n項(xiàng)和Sn.【解析】(1)設(shè)等差數(shù)列{an}的公差為d,等差數(shù)列{bn}的首項(xiàng)為b1,公比為q,依據(jù)已知條件可得?b1=2,d=2,q=2,∴an=2n-1,bn=2n.(2)由于{an}為等差數(shù)列,{bn}為等比數(shù)列,所以數(shù)列{anbn}的前n項(xiàng)和Sn可以表示為:Sn=1·2+3·22+5·23+…+(2n-1)·2n,①∴2Sn=1·22+3·23+…+(2n-3)·2n+(2n-1)·2n+1,②①-②得:-Sn=1·2+2·22+2·2