2022年秋高中數(shù)學(xué)第四章數(shù)列4.1數(shù)列的概念第2課時數(shù)列的遞推公式課后習(xí)題新人教A版選擇性必修第二冊

歡迎閱讀本文檔，希望本文檔能對您有所幫助！歡迎閱讀本文檔，希望本文檔能對您有所幫助！歡迎閱讀本文檔，希望本文檔能對您有所幫助！歡迎閱讀本文檔，希望本文檔能對您有所幫助！歡迎閱讀本文檔，希望本文檔能對您有所幫助！歡迎閱讀本文檔，希望本文檔能對您有所幫助！第2課時數(shù)列的遞推公式必備知識基礎(chǔ)練1.已知在數(shù)列{an}中,a1=2,an+1=an+n(n∈N*),則a4的值為()A.5 B.6 C.7 D.82.(2021河北承德一中高二月考)已知數(shù)列{an}的前n項(xiàng)和Sn=1n,則a6的值等于()A.120 B.-120 C.130 D3.(2021河南中原名校高三聯(lián)考)已知數(shù)列{an}的前n項(xiàng)和Sn=4n2-10n,則a2a6=()A.52 B.68 C.96 D.1084.已知數(shù)列{an}的前n項(xiàng)和為Sn,且Sn=2an-2,則a2=()A.4 B.2 C.1 D.-25.已知a1=1,an=n(an+1-an)(n∈N*),則數(shù)列{an}的通項(xiàng)公式是()A.2n-1 B.nC.n2 D.n6.若數(shù)列{an}滿足an+1=2an-1,且a8=16,則a6=.

7.在數(shù)列{an}中,a1=2,an+1=an+ln1+1n,求{an關(guān)鍵能力提升練8.已知數(shù)列{an},a1=2,a2=1,an+2=3an+1-an,則a6+a4-3a5的值為()A.3 B.-2 C.-1 D.09.已知數(shù)列{an},an+1=11-an,a1=3,則a2022=A.23 B.3 C.-12 D10.在數(shù)列{an}中,a1=1,對所有的n≥2,都有a1a2a3…an=n2,則a3+a5等于()A.259 B.2516 C.6116 11.(多選題)在無窮數(shù)列{an}中,若ap=aq(p,q∈N*),總有ap+1=aq+1,此時定義{an}為“階梯數(shù)列”.設(shè){an}為“階梯數(shù)列”,且a1=a4=1,a5=3,a8a9=23,則()A.a7=1 B.a8=2a4C.S10=10+33 D.a2020=112.設(shè){an}是首項(xiàng)為1的正項(xiàng)數(shù)列,且(n+1)an+12-nan2+an+1an=0(n∈N*),則它的通項(xiàng)公式a13.(2021陜西西安部分學(xué)校高二期末)已知數(shù)列{an}滿足a1=3,a2=8,an+2等于an+1an的個位數(shù),則a4=.

14.已知數(shù)列{an}滿足an+1=2若a1=67,試求a2021+a2022學(xué)科素養(yǎng)創(chuàng)新練15.(2021河南南陽高二期中)在數(shù)列{an}中,a1=35,an+1=an+1,0<aA.15 B.35 C.65 16.已知數(shù)列a1=1,a2,a3,…,an(n∈N*)的法則如下:若an為自然數(shù),則an+1=an-2,否則an+1=an+3,則a6=.

參考答案第2課時數(shù)列的遞推公式1.D因?yàn)閍1=2,an+1=an+n,所以a2=a1+1=2+1=3,a3=a2+2=3+2=5,a4=a3+3=5+3=8.2.Da6=S6-S5=16?15=-13.B由題意,可得當(dāng)n≥2時,an=Sn-Sn-1=4n2-10n-[4(n-1)2-10(n-1)]=8n-14,所以a2a6=(8×2-14)×(8×6-14)=68.4.ASn=2an-2?a1=S1=2a1-2?a1=2?a1+a2=S2=2a2-2?a2=4.故選A.5.D(方法一構(gòu)造法)由已知整理,得(n+1)an=nan+1,∴an+1n+1=a且ann=a11=(方法二累乘法)當(dāng)n≥2時,an…a3兩邊分別相乘,得ana1=n.∵a1=1,∴a6.194∵an+1=2an-1,∴a8=2a7-1=解得a7=172,又a7=2a6-1=172,解得a6=7.解由題意,得an+1-an=lnn+1∴an-an-1=lnnn-1(an-1-an-2=lnn-…,a2-a1=ln21∴當(dāng)n≥2時,an-a1=lnnn-1·n-1n-2∴an=2+lnn(n≥2).當(dāng)n=1時,a1=2+ln1=2,符合上式,∴an=2+lnn(n∈N*).8.D∵an+2=3an+1-an,∴an+2+an=3an+1.令n=4,得a6+a4=3a5,∴a6+a4-3a5=0.9.A由題意,可知:a1=3,a2=11-aa3=11a4=11-a5=11-a….∴數(shù)列{an}是一個以3為最小正周期的周期數(shù)列.∵2022÷3=674,∴a2022=a3=2310.C由題意a1a2=22,a1a2a3=32,a1a2a3a4=42,a1a2a3a4a5=52,則a3=3222=94,a5=524211.ACD因?yàn)閧an}為“階梯數(shù)列”,由a1=a4=1可得a2=a5,a3=a6,a4=a7,a5=a8,a6=a9,…,觀察可得a1=a4=a7=…=a3n-2=1(n∈N*),a2=a5=a8=…=a3n-1=3(n∈N*),a3=a6=a9=…=a3n(n∈N*),即數(shù)列{an}是以3為周期的周期數(shù)列,所以a7=1,a8=3,故A正確,B錯誤;a9=23a8=2,S10=(a1+a4+a7+a10)+(a2+a5+a8)+(a3+a6+a9)=10+33,故a2020=a1+3×673=a1=1,故D正確.故選ACD.12.1n把(n+1)an+12-nan2+an+1得[(n+1)an+1-nan](an+1+an)=0.∵an>0,∴an+1+an>0,∴(n+1)an+1-nan=0,∴an∴a2a1·a3a2·a4a3·…又a1=1,∴an=1na1=1又a1=1也適合上式,∴an=1n,n∈N*13.2由已知an+2等于an+1an的個位數(shù),又a1=3,a2=8,則a1a2=24,∴a3=4,則a2a3=32,∴a4=2.14.解∵a1=67∴a2=2a1-1=57∴a3=2a2-1=37∈0,12,∴a4=∴數(shù)列{an}是周期數(shù)列,且周期為3.∴a2021+a2022=a673×