全品高考备战2027年数学一轮学生用书04第39讲数列求和【答案】作业手册

第39讲数列求和1.C[解析]设等差数列{an}的公差为d.依题意可得,a2+a6=a1+d+a1+5d=10,即a1+3d=5①,又a4a8=(a1+3d)(a1+7d)=45,所以a1+7d=9②.由①②可得d=1,a1=2,所以S5=5a1+5×42×d=5×2+10=20.故选C2.A[解析]因为an=(-1)n-1(4n-3),所以S6=(a1+a2)+(a3+a4)+(a5+a6)=(1-5)+(9-13)+(17-21)=3×(-4)=-12,故选A.3.B[解析]Sn=12+24+38+…+n2n=1×12+2×122+3×123+…+n·12n,则12Sn=1×122+2×123+3×124+…+n·12n+1,两式相减得12Sn=12+122+124.C[解析]设bn=a2n-1,cn=a2n,则b1=a1=2,c1=a2=1.由题可得,a2n+1-a2n-1=2,即bn+1-bn=2,所以{bn}是以2为首项,2为公差的等差数列,则bn=2+2(n-1)=2n.由题可得a2n+2=2a2n,即cn+1=2cn所以{cn}是以1为首项,2为公比的等比数列,则cn=1×2n-1=2n-1.所以{an}的前20项和S20=(b1+b2+…+b10)+(c1+c2+…+c10)=(2+4+…+20)+(1+2+…+29)=10×5.C[解析]设{an}的公差为d,则d=a3-a12=3-12=1,所以an=n,所以Sn=n(n+1)2,所以1Sn=2n(n+1)=2×1n-1n+1,所以T6.99[解析]因为an=1n+nn+1-n,所以Sn=(2-1)+(3-2)+…+(n+1-n)=n+1-1,由Sn=9,得n+17.nn+1[解析]由a1+2a2+3a3+…+nan=n,得a1+2a2+3a3+…+(n-1)an-1=n-1(n≥2),两式相减得nan=1(n≥2),所以an=1n(n≥2),又a1=1满足上式,所以an=1n,所以ann+1=1n(n+1)=1n-1n+1.设数列ann+1的前n项和为Tn8.24[解析]因为a1=3,an+1+an=2n-1,所以a1+a2=1,所以a2=-2.因为an+1+an=2n-1①,所以an+2+an+1=2(n+1)-1②,由②-①得an+2-an=2,所以{a2n-1}是以3为首项,2为公差的等差数列,{a2n}是以-2为首项,2为公差的等差数列,所以a2n-1=2n+1,a2n=2n-4,所以a7=2×4+1=9,a6=2×3-4=2,所以S7=a1+a2+a3+a4+a5+a6+a7=(a1+a3+a5+a7)+(a2+a4+a6)=4×(3+9)29.解:(1)在4Sn=3an+4中取n=1,得a1=4,由4Sn=3an+4,4Sn-1=3an-1即4an=3an-3an-1(n≥2),∴an=-3an-1(n≥2),∴{an}是以4为首项,-3为公比的等比数列,∴an=4×(-3)n-1.(2)方法一:由(1)知bn=4n·3n-1,则Tn=4×(1+2×3+3×32+…+n×3n-1)①,则3Tn=4×[1×3+2×32+3×33+…+(n-1)×3n-1+n×3n]②,②-①得2Tn=4n·3n-4(1+3+32+…+3n-1)=4n·3n-4×3n-12=(4n-2)∴Tn=(2n-1)·3n+1.方法二:由(1)知bn=(-1)n-1nan=4n·3n-1,∴当n≥2时,Tn=Tn-1+4n·3n-1,两边同时减去(2n-1)·3n可得Tn-(2n-1)·3n=Tn-1-(2n-3)·3n-1(n≥2),故{Tn-(2n-1)·3n}为常数列,则Tn-(2n-1)·3n=1,可得Tn=(2n-1)·3n+1.10.A[解析]由题意,得S7=(a1+b1)+(a2+a3+b2+b3)+(a4+a5+b4+b5)+(a6+a7+b6+b7)=2+(2×2+22)+(2×4+24)+(2×6+26)=2+8+24+76=110.故选A.11.A[解析]数列{2n-1}是以1为首项,2为公差的等差数列,即1,3,5,7,9,11,13,…,数列{3n-2}是以1为首项,3为公差的等差数列,即1,4,7,10,13,16,19,…,所以数列{2n-1}与{3n-2}的公共项所构成的新数列{an}是以1为首项,6为公差的等差数列,所以an=6n-5,所以∑k=1101aka12.ABD[解析]对于A,由a1=0,因为an+1=an+(3)n+1,n为奇数,3an,n为偶数,所以a2=a1+3=3,a3=3a2=9,故A正确;对于B,当n≥2,n∈N*时,a2n=a(2n-1)+1=a2n-1+(3)2n=a2n-1+3n=a(2n-2)+1+3n=3a2n-2+3n(*),因为bn=a2n,所以bn-1=a2(n-1),故由(*)可得bn=3bn-1+3n,则bn3n=bn-13n-1+1,即数列bn3n是公差为1的等差数列,则bn3n=13b1+n-1=a23+n-1=n,可得bn=n·3n,故B正确;对于C,由Tn=1×3+2×32+3×33+…+n·3n,可得3Tn=1×32+2×33+3×34+…+n·3n+1,上面两式相减可得-2Tn=3+32+33+…+3n-n·3n+1=3(1-3n)1-3-n·3n+1,可得Tn=(2n-1)·3n+1+34,故C错误;对于D,由a2n=n·3n,a2n=a2n-1+3n,可得a2n-1=a2n-3n,则S213.ABD[解析]∵a1+2a2+…+2n-1an=n·2n+1,∴当n=1时,a1=4,当n≥2时,a1+2a2+…+2n-2an-1=(n-1)·2n,两式相减得2n-1an=n·2n+1-(n-1)·2n=(n+1)·2n,则an=2(n+1),又a1=4满足上式,∴an=2(n+1),故A正确;∵an=2(n+1),∴数列{an}是首项为4,公差为2的等差数列,∴{an}的前n项和Sn=4n+n(n-1)2×2=n2+3n,故B正确;{(-1)nan}的前100项和为-4+6-8+10+…-200+202=2+2+…+2=2×50=100,故C错误;{|an-10|}的前20项和为|a1-10|+|a2-10|+|a3-10|+…+|a20-10|=(10-a1)+(10-a2)+(10-a3)+(10-a4)+(a5-10)+(a6-10)+(a7-10)+…+(a20-10)=S20-214.-20272026[解析]由题意得an=(-1)n1n+1n+1,所以S2025=a1+a2+a3+…+a2024+a2025=-11-12+115.2027[解析]设等差数列{an}的公差为d,由1am-1am=1d1am-1-1am,可得1a1a2+1a2a16.解:(1)依题意得Snn=3n-2,即Sn=3n2-2n.当n≥2时,an=Sn-Sn-1=(3n2-2n)-[3(n-1)2-2(n-1)]=6当n=1时,a1=S1=3×12-2×1=1=6×1-5=1,满足上式.所以an=6n-5(n∈N+).(2)由(1)得bn=3anan+1=3(6n-5因为Tn<m20对所有n∈N+都成立,所以121-16n+1所以12≤m20,解得m≥10,故满足要求的最小正整数m17.解:(1)依题意,对任意的n≥2,都有an+1+an-1=2(an+1),故对任意的n≥2,an+1-an=an-an-1+2,所以对任意的n≥2,bn=bn-1+2,即bn-bn-1=2,所以数列{bn}是公差为2的等差数列,由a1=1,a3=9,得b1=a2-1,b2=9-a2,所以(9-a2)-(a2-1)=2,解得a2=4,所以b1=a2-1=3,所以bn=3+(n-1)×2=2n+