新教材2023高中數(shù)學第四章數(shù)列4.1數(shù)列的概念第2課時數(shù)列的遞推公式與前n項和分層演練新人教A版選擇性必修第二冊

4.1數(shù)列的概念第2課時數(shù)列的遞推公式與前n項和A級基礎鞏固1.已知數(shù)列{an}的首項為a1=1,若an+1=12an+12n,則此數(shù)列的第4項是(A.1 B.12 C.34解析:根據(jù)遞推公式結(jié)合a1=1逐個求解即可.答案:B2.在數(shù)列{an}中,a1=2,若an+1=an+ln1+1n,則an=(A.2+lnn B.2+(n-1)lnnC.2+nlnn D.1+n+lnn解析:由題意,可知an+1=an+lnn+即an+1-an=ln(n+1)-lnn,所以an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=lnn-ln(n-1)+ln(n-1)-ln(n-2)+…+ln2-ln1+2=2+lnn.答案:A3.已知數(shù)列{an}的前n項和為Sn,a1=45,若an+1=2an,0≤aA.408 B.410C.20485 D.解析:因為a1=45,an+1=所以a2=2a1-1=35,a3=2a2-1=1a4=2a3=25,a5=2a4=4所以各項以4為周期重復出現(xiàn).因為a1+a2+a3+a4=45+35+15+25=2,820所以S820=205×2=410.答案:B4.數(shù)列{an}滿足an+2=an+1+2an,且a1=1,a2=2,則a6=32.5.根據(jù)下列條件,寫出各數(shù)列的前4項,并歸納猜想數(shù)列的通項公式.(1)a1=0,an+1=an+2n-1(n∈N*);(2)a1=1,an+1=an+ann+1(n(3)a1=2,a2=3,an+2=3an+1-2an(n∈N*).解:(1)a1=0,a2=1,a3=4,a4=9.猜想an=(n-1)2.(2)a1=1,a2=32,a3=2,a4=52.猜想an=(3)a1=2,a2=3,a3=5,a4=9.猜想an=2n-1+1.6.在數(shù)列{an}中,a1=12,an=1-1an-1(n≥2,且n(1)求證:an+3=an;(2)求a1120.(1)證明:由題意,知an+3=1-1an+2=1-11-1an+1=1-11-11-1an=1-11-anan(2)解:由(1)知,數(shù)列{an}的周期T=3.因為1120=373×3+1,所以a1120=a1=12B級拓展提高7.已知數(shù)列{an}滿足a1=2,an=1-1an+1,記{an}的前n項之積為Tn,則T2022=A.-2 B.-1 C.1 D.2解析:由an=1-1an+1,得an+所以an+2=11-an+1=11-11-an=1-an1所以數(shù)列{an}是周期為3的周期數(shù)列.因為a1=2,所以a2=11-a1=11-2=-1,a3所以a1a2a3=2×(-1)×12=-1又因為2022=674×3,所以T2022=(a1a2a3)674答案:C8.已知數(shù)列{an}的通項公式an=ncosnπ2,設前n項和為Sn,則S1021=(A.1021 B.1020 C.510 D.1008解析:因為數(shù)列an=ncosnπ2呈周期性變化,觀察此數(shù)列規(guī)律如下:a1=0,a2=-2,a3=0,a4=4,a5=0,a6=-6,a7=0,a8=8,…,a1021故a1+a2+a3+a4=2,a5+a6+a7+a8=2,所以S1021=(a1+a2+a3+a4)+…+(a1017+a1018+a1019+a1020)+a1021=10204×2+a1021=510+0=510答案:C9.已知數(shù)列{an}滿足a1=1,若當n≥2時,n2(an-an-1)+an-1=0,則an=n+解析:因為當n≥2時,n2(an-an-1)+an-1=0,所以nan(當n≥2時,nan=n+1n(n-1)an-1=n+1n·nn-1(n-2)·an-2=…=n+1n·nn所以an=n+因為當n=1時,n+12n=1=a1,所以a10.設{an}是首項為1的正項數(shù)列,且(n+1)·an+12-nan2+an+1an=0(n∈N*),求數(shù)列解:因為(n+1)an+12-nan2+a所以(an+1+an)[(n+1)an+1-nan]=0.又因為an>0,所以an+1+an>0.所以(n+1)an+1-nan=0,即an+1所以當n≥2時,an=anan-1·an-1an-2·an-2an-3·…·a3a2又因為a1=1也適合此式,所以an=1n11.已知數(shù)列{an}的前n項和為Sn,a1=1,且Sn=12an(n+1)(1)求a2,a3;(2)求數(shù)列{an}的通項公式.解:(1)因為a1=1,且Sn=12an(n+所以當n=2時,1+a2=12×3a2,解得a2=當n=3時,1+2+a3=12×4×a3,解得a3=3(2)當n≥2時,an=Sn-Sn-1=12an(n+1)-12an-1·整理,得ann=因為a2=2,所以ann=an-1n-1=…=a因為當n=1時也成立,所以an=n.C級挑戰(zhàn)創(chuàng)新12.多空題設數(shù)列{an}的前n項和為Sn,若Sn=n2-4n,則a1=-3,a4+a5=8.解析:因為Sn=n2-4n,所以當n=1時,a1=S1=-3.a4+a5=S5-S3=8.13.多空題若數(shù)列{an}的前n項和為Sn=n2+n+1,bn=(-1)n(an-2)(n∈N*),則數(shù)列{an}的通項公式為an=3,n=1,2n,n≥2解析:由題可知,當n≥2時,an=Sn-Sn-1=n2+n+1-(n-1)2-(n-1)-1=2n.當n=1時,a1=S1=12+1+1=3.所以an=3因為bn=(-1)n(an-2),所以bn=-所以數(shù)列{bn}的前50項和為-1+2×1-2×2+2×3-2×4+…+2×49=-1+2×(1-2+3-4+…+49)=-1+2×25=49.14.多空題定義“等積數(shù)列”:如果一個數(shù)列從第2項起,每一項與它的前一項的乘積都等于同一個不為零的常數(shù),那么這個數(shù)列叫做等積數(shù)列,這個常數(shù)叫做等積數(shù)列的公積.已知數(shù)列{an}是首項a1=2,公積為-6的等積數(shù)列,則a3=2,數(shù)列{an}的前n項和Sn=-解析:因為數(shù)列{an}是等積數(shù)列,a1=2,公積為-6,所以a2=-3,a3=2,a4=-3……所以前