2019版高考數(shù)學(xué)二輪復(fù)習(xí)專(zhuān)題四數(shù)列專(zhuān)題對(duì)點(diǎn)練14數(shù)列與數(shù)列不等式的證明及數(shù)列中的存在性問(wèn)題文

歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助!歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助!歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助!歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助!歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助!歡迎閱讀本文檔，希望本文檔能對(duì)您有所幫助!專(zhuān)題對(duì)點(diǎn)練14數(shù)列與數(shù)列不等式的證明及數(shù)列中的存在性問(wèn)題1.已知等比數(shù)列{an},a1=,公比q=.(1)Sn為{an}的前n項(xiàng)和,證明:Sn=1-(2)設(shè)bn=log3a1+log3a2+…+log3an,求數(shù)列{bn2.已知數(shù)列{an}滿足a1=3,an+1=3a(1)證明:數(shù)列1an-1(2)令bn=a1a2…an,求數(shù)列1bn的前n項(xiàng)和3.已知數(shù)列{an}的前n項(xiàng)和Sn=1+λan,其中λ≠0.(1)證明{an}是等比數(shù)列,并求其通項(xiàng)公式;(2)若S5=3132,求λ的值4.在數(shù)列{an}中,設(shè)f(n)=an,且f(n)滿足f(n+1)-2f(n)=2n(n∈N*),且a1=(1)設(shè)bn=an2n-(2)求數(shù)列{an}的前n項(xiàng)和Sn.5.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且(3-m)Sn+2man=m+3(n∈N*),其中m為常數(shù),且m≠-3.(1)求證:{an}是等比數(shù)列;(2)若數(shù)列{an}的公比q=f(m),數(shù)列{bn}滿足b1=a1,bn=f(bn-1)(n∈N*,n≥2),求證:1bn為等差數(shù)列,并求b6.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=-2,且滿足Sn=an+1+n+1(n∈N*).(1)求數(shù)列{an}的通項(xiàng)公式;(2)若bn=log3(-an+1),求數(shù)列1bnbn+2的前n項(xiàng)和T7.(2018天津模擬)已知正項(xiàng)數(shù)列{an},a1=1,a2=2,前n項(xiàng)和為Sn,且滿足Sn+1Sn-1+Sn-(1)求數(shù)列{an}的通項(xiàng)公式;(2)記cn=1Sn·Sn+1,數(shù)列{cn}的前n項(xiàng)和為8.已知數(shù)列{an}的前n項(xiàng)和為Sn,a1=2,2Sn=(n+1)2an-n2an+1,數(shù)列{bn}滿足b1=1,bnbn+1=λ·2a(1)求數(shù)列{an}的通項(xiàng)公式;(2)是否存在正實(shí)數(shù)λ,使得{bn}為等比數(shù)列?并說(shuō)明理由.

專(zhuān)題對(duì)點(diǎn)練14答案1.(1)證明因?yàn)閍n=13×13n-所以Sn=1-(2)解bn=log3a1+log3a2+…+log3an=-(1+2+…+n)=-所以{bn}的通項(xiàng)公式為bn=-n(2.解(1)∵an+1=3an-1an+1,∴an+1-1∴1a∴1a∵a1=3,∴1a∴數(shù)列1an-1是以為首項(xiàng),為公差的等差數(shù)列,∴1an-1=12+1(2)∵bn=a1a2…an∴bn=31×42×∴1bn=2∴Sn=212-13+13-3.解(1)由題意得a1=S1=1+λa1,故λ≠1,a1=11-λ,a由Sn=1+λan,Sn+1=1+λan+1得an+1=λan+1-λan,即an+1(λ-1)=λan.由a1≠0,λ≠0得an≠0,所以an因此{(lán)an}是首項(xiàng)為11-λ于是an=11(2)由(1)得Sn=1-λλ由S5=3132得1-λ即λλ解得λ=-1.4.(1)證明由已知得an+1=2an+2n,∴bn+1=an+12n=2∴bn+1-bn=1.又a1=1,∴b1=1,∴{bn}是首項(xiàng)為1,公差為1的等差數(shù)列.(2)解由(1)知,bn=an2∴an=n·2n-1.∴Sn=1+2×21+3×22+…+n·2n-1,2Sn=1×21+2×22+…+(n-1)·2n-1+n·2n,兩式相減得-Sn=1+21+22+…+2n-1-n·2n=2n-1-n·2n=(1-n)2n-1,∴Sn=(n-1)·2n+1.5.證明(1)由(3-m)Sn+2man=m+3,得(3-m)Sn+1+2man+1=m+3,兩式相減,得(3+m)an+1=2man.∵m≠-3,∴an∴{an}是等比數(shù)列.(2)由(3-m)Sn+2man=m+3,得(3-m)S1+2ma1=m+3,即a1=1,∴b1=1.∵數(shù)列{an}的公比q=f(m)=2m∴當(dāng)n≥2時(shí),bn=f(bn-1)=32∴bnbn-1+3bn=3bn-1,∴1b∴1bn是以1為首項(xiàng),為公差的等差數(shù)列,∴1bn=又1b1=1也符合,∴bn=6.(1)解∵Sn=an+1+n+1(n∈N*),∴當(dāng)n=1時(shí),-2=a2+2,解得a2=-8.當(dāng)n≥2時(shí),an=Sn-Sn-1=an+1+n+1-12即an+1=3an-2,可得an+1-1=3(an-1).當(dāng)n=1時(shí),a2-1=3(a1-1)=-9,∴數(shù)列{an-1}是等比數(shù)列,首項(xiàng)為-3,公比為3.∴an-1=-3n,即an=1-3n.(2)證明bn=log3(-an+1)=n,∴1b∴Tn=121-13+12-7.(1)解由Sn+1Sn-1+Sn-1Sn+1=4Sn2Sn+1Sn即(Sn+1+Sn-1)2=(2Sn)2.由數(shù)列{an}的各項(xiàng)均為正數(shù),得Sn+1+Sn-1=2Sn,所以數(shù)列{Sn}為等差數(shù)列.由a1=1,a2=2,得S1=a1=1,S2=a1+a2=3,則數(shù)列{Sn}的公差為d=S2-S1=2,所以Sn=1+(n-1)×2=2n-1.當(dāng)n≥2時(shí),an=Sn-Sn-1=(2n-1)-(2n-3)=2,而a1=1不適合上式,所以數(shù)列{an}的通項(xiàng)公式為an=1(2)證明由(1)得cn=1S則Tn=c1+c2+c3+…+cn=121-13+13又Tn=121-12n+1是關(guān)于n的增函數(shù),則Tn≥8.解(1)由2Sn=(n+1)2an-n2an+1,得2Sn-1=n2an-1-(n-1)2an,∴2an=(n+1)2an-n2an+1-n2an-1+(n-1)2an,∴2an=an+1+an-1,∴數(shù)列{an}為等差數(shù)列.∵2S1=(1+1)2a1-a2,∴4=8-a2∴a2=4.∴d=a2-a1=4-2=2.∴an=2+2(n-1)=2n.(