【全程復(fù)習(xí)方略】2020年人教A版數(shù)學(xué)理(廣東用)課時(shí)作業(yè)：第五章-第三節(jié)等比數(shù)列及其前n項(xiàng)和

溫馨提示：此套題為Word版，請(qǐng)按住Ctrl,滑動(dòng)鼠標(biāo)滾軸，調(diào)整合適的觀看比例，答案解析附后。關(guān)閉Word文檔返回原板塊。課時(shí)提升作業(yè)(三十二)一、選擇題1.已知等比數(shù)列{an}的公比q=2,其前4項(xiàng)和S4=60,則a2等于()(A)8 (B)6 (C)-8 (D)-62.等比數(shù)列{an}中,若log2(a2a98)=4,則a40a(A)-16 (B)10 (C)16 (D)2563.在正項(xiàng)等比數(shù)列{an}中,a1,a19分別是方程x2-10x+16=0的兩根,則a8·a10·a12等于()(A)16 (B)32 (C)64 (D)2564.(2021·汕頭模擬)等比數(shù)列{an}中,a1=3,a4=24,則a3+a4+a5=()(A)84 (B)83 (C)82 (D)815.(2021·沈陽(yáng)模擬)已知數(shù)列{an}滿足log3an+1=log3an+1(n∈N*)且a2+a4+a6=9,則loQUOTE(a5+a7+a9)的值是()(A)-5 (B)-QUOTE (C)5 (D)QUOTE6.設(shè)等比數(shù)列{an}的前n項(xiàng)和為Sn,若a2011=3S2010+2022,a2010=3S2009+2022,則公比q=()(A)4 (B)1或4(C)2 (D)1或27.(2021·東莞模擬)已知等差數(shù)列{an}的公差為2,若a1,a3,a4成等比數(shù)列,則a2=()(A)-4 (B)-6 (C)-8 (D)-108.(2021·漢中模擬)在等比數(shù)列{an}中,a6與a7的等差中項(xiàng)等于48,a4a5a6a7a8a9(A)5n-4 (B)4n-3(C)3n-2 (D)2n-1二、填空題9.(2022·廣東高考)若等比數(shù)列{an}滿足a2a4=QUOTE,則a1QUOTEa5=.10.(2021·佛山模擬)已知等比數(shù)列{an}的首項(xiàng)為2,公比為2,則QUOTE=.11.數(shù)列1QUOTE,2QUOTE,3QUOTE,4QUOTE,…的前n項(xiàng)和為.12.(力氣挑戰(zhàn)題)設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,已知a1=1,Sn+1=2Sn+n+1(n∈N*),則數(shù)列{an}的通項(xiàng)公式an=.三、解答題13.(2021·湛江模擬)已知等比數(shù)列{an}的前n項(xiàng)中,a1最小,且a1+an=66,a2an-1=128,前n項(xiàng)和Sn=126,求n和公比q.14.已知數(shù)列{an}中,a1=1,前n項(xiàng)和為Sn且Sn+1=QUOTESn+1,n∈N*.(1)求數(shù)列{an}的通項(xiàng)公式.(2)求數(shù)列{QUOTE}的前n項(xiàng)和Tn.15.(力氣挑戰(zhàn)題)設(shè)一元二次方程anx2-an+1x+1=0(n=1,2,3,…)有兩根α和β,且滿足6α-2αβ+6β=3.(1)試用an表示an+1.(2)求證:數(shù)列{an-QUOTE}是等比數(shù)列.(3)當(dāng)a1=QUOTE時(shí),求數(shù)列{an}的通項(xiàng)公式.答案解析1.【解析】選A.S4=60,q=2?QUOTE=60?a1=4,∴a2=a1q=4×2=8.2.【解析】選C.a40a60=a2a98,依據(jù)log2(a2a98)=4即可求解.依據(jù)已知a2a98=243.【解析】選C.依據(jù)根與系數(shù)的關(guān)系得a1a19=16,由此得a10=4,a8a12=16,故a8·a10·a4.【解析】選A.q3=QUOTE=QUOTE=8,∴q=2,a3=QUOTE=QUOTE=12,a5=a4·q=24×2=48,∴a3+a4+a5=12+24+48=84.5.【思路點(diǎn)撥】依據(jù)數(shù)列滿足log3an+1=log3an+1(n∈N*)且a2+a4+a6=9可以確定數(shù)列是公比為3的等比數(shù)列,再依據(jù)等比數(shù)列的通項(xiàng)公式即可通過a2+a4+a6=9求出a5+a7+a9的值.【解析】選A.由log3an+1=log3an+1(n∈N*),得an+1=3an,又由于an>0,所以數(shù)列{an}是公比為3的等比數(shù)列,a5+a7+a9=(a2+a4+a6)×33=35,所以loQUOTE(a5+a7+a9)=-log335=-5.6.【解析】選A.由a2011=3S2010+2022,a2010=3S2009+2022兩式相減得a2011-a2010=3a2010,即q=4.7.【解析】選B.a1=a2-2,a3=a2+2,a4=a2+4,∵a1,a3,a4成等比數(shù)列,∴QUOTE=a1·a4,∴(a2+2)2=(a2-2)(a2+4),QUOTE+4a2+4=QUOTE+2a2-8,∴2a2=-12,即a2=-6.8.【解析】選D.設(shè)等比數(shù)列{an}的公比為q,由a6與a7的等差中項(xiàng)等于48,得a6+a7=96,即a1q5(1+q)=96①.由等比數(shù)列的性質(zhì),得a4a10=a5a9=a6a8=QUOTE.由于a4a5a6a7a8a9a10=1286,則QUOTE=1286=(26)7,即a1q6=26②.由①②解得a1=1,q=2,∴Sn=QUOTE=2n-1,故選D.9.【思路點(diǎn)撥】本題考查了等比數(shù)列的性質(zhì):已知m,n,p∈N*,若m+n=2p,則am·an=QUOTE.【解析】∵a2a4=QUOTE,∴QUOTE=QUOTE,∴a1QUOTEa5=QUOTE=QUOTE.答案:QUOTE10.【解析】由題意知an=2n,所以QUOTE=QUOTE=QUOTE=22=4.答案:411.【解析】設(shè)所求的前n項(xiàng)和為Sn,則Sn=(1+2+…+n)+(QUOTE+QUOTE+…+QUOTE)=QUOTE+1-QUOTE.答案:QUOTE+1-QUOTE12.【解析】∵Sn+1=2Sn+n+1,當(dāng)n≥2時(shí)Sn=2Sn-1+n,兩式相減得:an+1=2an+1,∴an+1+1=2(an+1),即QUOTE=2.又S2=2S1+1+1,a1=S1=1,∴a2=3,∴QUOTE=2,∴{an+1}是首項(xiàng)為2,公比為2的等比數(shù)列,∴an+1=2n即an=2n-1(n∈N*).答案:2n-1【方法技巧】含Sn,an問題的求解策略當(dāng)已知含有Sn+1,Sn之間的等式時(shí),或者含有Sn,an的混合關(guān)系的等式時(shí),可以接受降級(jí)角標(biāo)或者升級(jí)角標(biāo)的方法再得出一個(gè)等式,兩個(gè)等式相減就把問題轉(zhuǎn)化為數(shù)列的通項(xiàng)之間的遞推關(guān)系式.13.【解析】由于{an}為等比數(shù)列,所以a1an=a2an-1,∴QUOTE且a1≤an,解得a1=2,an=64,依題意知q≠1,∵Sn=126,∴QUOTE=126?q=2,∵2qn-1=64,∴n=6.14.【解析】(1)由Sn+1=QUOTESn+1,得當(dāng)n≥2時(shí)Sn=QUOTESn-1+1,∴Sn+1-Sn=QUOTE(Sn-Sn-1),即an+1=QUOTEan,∴QUOTE=QUOTE,又a1=1,得S2=QUOTEa1+1=a1+a2,∴a2=QUOTE,∴QUOTE=QUOTE,∴數(shù)列{an}是首項(xiàng)為1,公比為QUOTE的等比數(shù)列,∴an=(QUOTE)n-1.(2)∵數(shù)列{an}是首項(xiàng)為1,公比為QUOTE的等比數(shù)列,∴數(shù)列{QUOTE}是首項(xiàng)為1,公比為QUOTE的等比數(shù)列,∴Tn=QUOTE=3[1-(QUOTE)n].15.【解析】(1)∵一元二次方程anx2-an+1x+1=0(n=1,2,3,…)有兩根α和β,由根與系數(shù)的關(guān)系易得α+β=QUOTE,αβ=QUOTE,∵6α-2αβ+6β=3,∴QUOTE-QUOTE=3,即an+1=QUOTEan+QUOTE.(2)∵an+1=QUOTEan+QUOTE,∴an+1-QUOTE=QUOTE(an-QUOTE),當(dāng)an-QUOTE≠0時(shí),QUOTE=QUOTE,當(dāng)an-QUOTE=0,即an=QUOTE時(shí),此時(shí)一元二次方程為QUOTEx2-QUOTEx+1=0,即2x2-2x+3=0,∵Δ=4-24<0,∴不合題意,即數(shù)列{an-QUOTE}是等比數(shù)列.(3)由(2)知:數(shù)列{an-QUOTE}是以a1-QUOTE=QUOTE-QUOTE=QUOTE為首項(xiàng),公比為QUOTE的等比數(shù)列,∴an-QUOTE=QUOTE×(QUOTE)n-1=(QUOTE)n,即an=(QUOTE)n+QUOTE,∴數(shù)列{an}的通項(xiàng)公式是an=(QUOTE)n+QUOTE.【變式備選】定義:若數(shù)列{An}滿足An+1=QUOTE,則稱數(shù)列{An}為“平方遞推數(shù)列”.已知數(shù)列{an}中,a1=2,點(diǎn)(an,an+1)在函數(shù)f(x)=2x2+2x的圖象上,其中n為正整數(shù).(1)證明:數(shù)列{2an+1}是“平方遞推數(shù)列”,且數(shù)列{lg(2an+1)}為等比數(shù)列.(2)設(shè)(1)中“平方遞推數(shù)列”的前n項(xiàng)之積為Tn,即Tn=(2a1+1)(2a2+1)…(2an+1),求數(shù)列{an}的通項(xiàng)公式及Tn關(guān)于n的表達(dá)式.【解析】(1)由條件得:an+1=2QUOTE+2an,∴2an+1+1=4QUOTE+4an+1=(2an+1)2,∴{2an+1}是“平方遞推數(shù)列”.∵