02 模块二　数列 【答案】听课手册

模块二数列微专题5等差数列、等比数列微点1例1(1)A[解析]设等差数列{an}的公差为d,由a2+a7=a8+1,得2a1+7d=a1+7d+1,可得a1=1,因为a2,a4,a8成等比数列,所以a42=a2a8,即(1+3d)2=(1+d)(1+7d),因为d≠0,所以d=1,所以a2023=a1+(2023-1)×d=2023.故选(2)解:①方法一:因为数列{bn}是以1为首项,3为公比的等比数列,所以bn=3n-1.因为(n+1)an=n2-8n+k,所以a1=k-72,a2=k-123因为数列{an}是等差数列,所以2a2=a1+a3,即2×k-123=k-72+k-154,解得k=-9,所以(n+1)an=n2-8n-9=(n+方法二:因为数列{bn}是以1为首项,3为公比的等比数列,所以bn=3n-1.设数列{an}的公差为d,则an=a1+(n-1)d=dn+a1-d.所以(n+1)an=(n+1)(dn+a1-d)=dn2+a1n+a1-d=n2-8n+k,所以d=1,a1=-8,a1-d=k②由①知cn=anbn当n≤8时,cn<0;当n=9时,cn=0;当n≥10时,cn>0.当n≥10时,cn+1-cn=n-83n-n-93n-1=所以数列{cn}的最大项是c10=13【自测题】1.A[解析]方法一:设等差数列{an}的公差为d,因为a2=-2,a1+a10=a3+4,所以a1+d=-2,2a1+9d=a1+2d+4,方法二:由等差数列的性质可得a10+a1-a3=a10-2d=a8=4,故a5=a2+a2.314[解析]设等比数列{an}的公比为q,由a3=1,2S3=7a2,可得a1q2=1,2a1+2=5a1q,解得a1=14,q=2或a1=4,q=12.当a1=14,q=2时,S5=a1(1-q5)1-q=14×1-251-2=314;当a1=4,微点2例2(1)C(2)A[解析](1)方法一:由题易知公比q≠1且q≠0,∵S6=21S2,∴1-q6=21(1-q2),∴q4+q2-20=0,∴q2=4,则S8S4=1+q4=17,∴S8=17×(-5)=-85.方法二:由等比数列的性质可得S2,S4-S2,S6-S4,S8-S6成等比数列,因此(S4-S2)2=S2(S6-S4),将S4=-5,S6=21S2代入上式,解得S2=-1或S2=54.当S2=54时,S4=S2(1+q2)>0,不满足题意,所以S2=-1,则S6=-21,由等比数列的性质可知(S6-S4)2=(S4-S2)(S8-S6),得S8=-85.故选(2)由已知得S3,S6-S3,S9-S6,即3,S6-3,18-S6成等差数列,所以2×(S6-3)=3+(18-S6),可得S6=9.故选A.【自测题】1.B[解析]设各项均为正数的等比数列{an}的公比为q,因为a685=a684+2a683,所以a1q684=a1q683+2a1q682,即q2-q-2=0,解得q=2或q=-1.因为数列{an}是各项均为正数的等比数列,所以q=2,所以an=a1·2n-1.因为aman=a1qm-1·a1qn-1=a12·2m+n-2=2a1,所以m+n-2=2,即m+n=4,且m,n∈N*,当m=1,n=3时,1m+4n=73;当m=n=2时,1m2.8[解析]因为S16=16(a1+a16)2=8(a1+a16)=8(a8+a9)>0,所以a8+a9>0,又a7+a9=2a8<0,所以a9>0,则公差d=a9-a8>0,所以{an}为递增的等差数列,即a1<a2<…<a8<0<a9<…,所以当微点3例3解:(1)由题可知an>1,因为an+1=an2-2an+2,所以an+1-1=(an-1)则ln(an+1-1)=ln(an-1)2=2ln(an-1),又ln(a1-1)=ln2,所以数列{ln(an-1)}是以ln2为首项,2为公比的等比数列,则ln(an-1)=2n-1·ln2=ln22n-1,所以an=(2)证明:由an+1=an2-2an+2,得an+1-2=an(an则1an+1-2所以1an=1an-2-2an+1-2,所以bn=1an所以Sn=b1+b2+…+bn=2a1-2-2a2-2+2a2-因为222n-1>0,所以2-222n【自测题】解:(1)a2=a12=1,a3=23a2+2=由题意得a2n+1+23=22n+1a2n+83=22n+1·a2n-122n-1+83又a1+23=83≠0,所以数列a(2)由(1)知bn=a2n-1=83·4n-1-2故Tn=83×(40+41+42+…+4n-1)-23n=83×1-4n1-4-2微点4例4ACD[解析]根据题意知,从上往下每一层的球的个数构成数列{an},分析可得a1=1,a2=3,a3=6,…,则a2-a1=3-1=2,a3-a2=6-3=3,a4-a3=4,…,归纳可得an+1-an=n+1,故B错误;an=(an-an-1)+(an-1-an-2)+…+(a2-a1)+a1=n+(n-1)+(n-2)+…+2+1=n(n+1)2,故C正确;a4=10,a5=15,则S5=a1+a2+a3+a4+a5=1+3+6+10+15=35,故A正确;由an=n(n+1)2,得1an=2n(n+1)=21n-1n+1,则1a1+1a2+1a3+…+1【自测题】2.6[解析]设蒲的长度构成等比数列{an},其前n项和为An,易知数列{an}的首项a1=3,公比为12,设莞的长度构成等比数列{bn},其前n项和为Bn,易知数列{bn}的首项b1=1,公比为2,则An=31-12n1-12,Bn=1-2n1-2,由31-12n1-12=1-2n1-1.C[解析]由题意,{an}为等比数列,且an+1=2Sn+2①,∴an=2Sn-1+2(n≥2)②,①-②得an+1-an=2an(n≥2),即an+1=3an(n≥2),故数列{an}是公比为3的等比数列.又∵a2=2(a1+1)=3a1,∴a1=2,故a4=2×33=54.2.C[解析]取m=1,则an+1=a1an=2an,故数列{an}是公比为2的等比数列,又a1=2,所以an=2n,所以ak+1+ak+2+…+ak+10=2k+1(1-210)1-2=2k+11-2k+13.D[解析]设OD1=DC1=CB1=BA1=1,则DD1=0.5,CC1=k1,BB1=k2,AA1=k3.由题意知,点A的坐标为(4,0.5+k1+k2+k3),即(4,3k3+0.2),所以kOA=3k3+0.24=0.725,所以k3=4.C[解析]方法一:若{an}为等差数列,则设数列{an}的公差为d,可得Sn=na1+n(n-1)2d,则Snn=a1+(n-1)2d,所以Snn为首项为a1,公差为d2的等差数列,即甲是乙的充分条件.反之,若Snn为等差数列,则可设Snn的公差为D,故Snn=S1+(n-1)D,即Sn=nS1+n(n-1)D=na1+n(n-1)D,利用公式an=S1,n=1,Sn-Sn-1,n≥2,可得a方法二:若{an}为等差数列,设数列{an}的公差为d,则Sn=na1+n(n-1)2d,Snn=a1+d2(n-1),所以Snn为等差数列,其首项为a1,公差为d2,即甲是乙的充分条件.反之,若Snn为等差数列,可设数列Snn的公差为D,则Sn+1n+1-Snn=D,Snn=S1+(n-1)D,即Sn=nS1+n(n-1)D,Sn-1=(n-1)S1+(n-1)(n-2)D(n≥2),当n≥2时,上述两式相减得an=Sn-Sn-1=S1+2(n-1)D,当n=1时,上式成立,故an=a1+2(n-1)D,又an+1-an=a1+2nD-[a1+2(n-1)D]=5.48384[解析]方法一:设数列{an}(1≤n≤9,n∈N*)的前3项的公差为d,后7项的公比为q,q>0,则q4=a9a5=19212=16,可得q=2,则a3=1+2d=a5q2,即1+2d=3,可得d=1.易得a3=3,a7=a3q4=48,数列{an}的所有项的和为a1+a2+…+a9=1+2+3+3×2+…+3×26方法二:因为{an}(1≤n≤9,n∈N*)的后7项成等比数列,所以a72=a5a9=12×192=482,且an>0,所以a7=48.又因为a52=a3a7,则a3=a52a7=3.设后7项的公比为q,q>0,则q2=a5a3=4,可得q=2,所以a1+a2+a3=3(a1+a3)2=6,a3+a4+a5+a6+a7+a8+a9=a3-a9q1-q6.-2[解析]设等比数列{an}的公比为q(q≠0),由a2a4a5=a3a6,结合等比数列的性质有a4a5=a3a6,得a2=1,所以a9a10=a2q7a2q8=a22q15=q15=-8,则q5=-2,故a7=a2q5=-

微专题6递推数列与数列求和微点1例1(1)D(2)14[解析](1)由Sn=Sn+1-3an-2,得Sn+1-Sn=3an+2,即an+1=3an+2,即an+1+1=3(an+1),又a1+1=3,所以数列{an+1}是以3为首项,3为公比的等比数列,所以an+1=3n,即an=3n-1,所以S20=3+32+…+320-20=3×(1-320)1-3(2)∵anan+1=an-1,∴an+1=1-1an,又a1=-3,∴a2=43,a3=14,a4=-3,a5=43,…,∴{an}是以3为周期的数列,则a105【自测题】1.B[解析]由a1+a2n-1=4n-6,令n=1,得a1+a1=4-6=-2,则a1=-1,令n=4,则a1+a7=4×4-6=10,则a7=11.故选B.2.11220232024[解析]由a1=12,可得1a1=2,∵1an+1-1an=2n+2,∴1a2-1a1=4,1a3-1a2=6,…,1an-1an-1=2n(n≥2),各式相加,可得1an=2+4+6+…+2n=n(2+2n)2=n(n+1)(n≥2),则an=1n(n+1)=1n-1n+1(n≥2),又a1=12满足上式,∴an=1微点2例2解:(1)当n=1时,a1=S1=12+12+1当n≥2时,an=Sn-Sn-1=12n2+12n+1-12(n-1)2-12(n-综上,an=2(2)因为b1a2+b2a3+…+bnan所以当n=1时,b1a2=92-32=3,所以b1=6.当n≥2时,由b1a2+b2a3+…+bnan+1=12×3n+1-32,又当n=1时,b1=6=(1+1)×31满足上式,所以bn=(n+1)·3n.所以Tn=b1+b2+…+bn=2×3+3×32+…+(n+1)·3n,3Tn=2×32+3×33+…+(n+1)·3n+1,所以-2Tn=2×3+32+33+…+3n-(n+1)·3n+1=6+9(1-3n-1)1-3-(n+1)·3所以Tn=2n+14×3n+1【自测题】解:(1)由题知an>0.∵an=Sn-Sn-1(n≥2),∴an=(Sn-Sn-1)(Sn+又an=Sn+Sn-1(n≥2),∴Sn-S∴数列Sn是以S1=a1=1=1为首项,1为公差的等差数列,∴Sn=1+(n-1)=n,∴S当n≥2时,an=Sn-Sn-1=n2-(n-1)2=2n-1,当n=1时,a1=1,满足上式,∴数列{an}的通项公式为an=2n-1.(2)证明:由(1)可知,an=2n-1,则an+22nana故Tn=11×20-13×21+13×21-15×2∵1(2n+1)2n>微点3例310×3k-1[解析]设数列Ak中,0的个数为ak,1的个数为bk,则ak+1=ak+2bk①,bk+1=2ak+bk②,①+②得ak+1+bk+1=3(ak+bk),又a1+b1=5,∴数列{ak+bk}是以5为首项,3为公比的等比数列,∴ak+bk=5×3k-1.①-②得,ak+1-bk+1=-(ak-bk),又a1-b1=-1,∴数列{ak-bk}是以-1为首项,-1为公比的等比数列,∴ak-bk=(-1)k,∴ak=5×3k-1+(-1)k2,bk=5×3k-1-(-1)k2,∴Sk=0×ak+1×bk=bk,【自测题】1.ACD[解析]对于A,由题可知,a3=2,a4=5,a5=12,a6=29,a7=70,a8=169,a9=408,a10=985,故A正确.对于B,因为a2-a1=1,a3-a2=1,a4-a3=3≠a3-a2,故B错误.对于C,由an+2=2an+1+an可设数列{an+1+kan}是公比为q的等比数列,则an+2+kan+1=q(an+1+kan),所以an+2=(q-k)an+1+qkan,所以q-k=2,qk=1,所以k=-1+2,q=1+2或k=-1-2,q=1-2.当k=-1+2,q=1+2时,an+1+(-1+2)an=1×(1+2)n-1,当k=-1-2,q=1-2时,an+1+(-1-2)a2+1-(-2+1)×(-3+22)n1-(-3+22)n=(2+1)×1-(-3+22)n+11-(-3+22)n,-3+22∈(2.59[解析]将已知数列分组,每组的第一项均为1,即第一组:1;第二组:1,3;第三组:1,3,5.以此类推,将各组数据之和记为数列{bm},则bm=m(1+2m-1)2=m2,记数列{bm}的前m项和为Tm,则Tm=12+22+…+m2=m(m+1)(2m+1)6,∴T10=10×11×216=385<400,T11=11×12×236=506>400.∵b1+b2+…+b10对应{an}中的第1+2+3+…+10=10×112=55项,即S55=T10,∴S58=385+1+3+5=394<400,S59=385+1+1.52403-n+32n[解析]由题意知,对折方式分为横折和竖折两种,第一次对折时,有横、竖两种折法,可得到2种规格的图形,第二次对折时,第一个图形可得到2种规格的图形,从第二个图形开始,对折得到的2种图形中有一种与前面得到的图形相同,故对折2次共可以得到3种不同规格的图形,对折3次共可以得到4种不同规格的图形,对折4次共可以得到5种不同规格的图形……对折n次共可以得到(n+1)种不同规格的图形.在每次对折中所得不同规格的图形的面积相等,对折n次得到的图形中有一种是横折n次得到的,其面积为12nS,其中S=20×12=240(dm2),故对折n次得到的(n+1)种规格的图形的面积之和Sn=n+12nS,所以∑k=1nSk=221S+322S+…+n+12nS.设Tn=221+322+…+n+12n,则12Tn=222+323+…+n+12n+1,两式相减,得12Tn=1+122+123+…2.解:(1)因为S1=a1=1,所以S1a所以数列Snan是首项为1,公差为所以Snan=1+(n-1)·13=n+23,所以S当n≥2时,an=Sn-Sn-1=n+23an-n+13所以(n-1)an=(n+1)an-1,即anan则an=anan-1×an-1an-2×…×a3a2×a2a又a1=1满足上式,所以{an}的通项公式为an=n((2)证明:1a1+1a2+…+1an=2×11×2+12×3+

微专题7数列中的双数列问题微点1例1解:(1)设数列{an}的公差为d,∵T3=b1+b2+b3=a1-6+2a2+a3-6=4a2-12=16,∴a2=a1+d=7,又S4=4a1+4×32d=32,即a1+32d=8,∴a1=5,∴an=a1+(n-1)d=2n+3.(2)证明:由(1)可得Sn=na1+n(n-1)当n为偶数时,Tn=(b1+b3+…+bn-1)+(b2+b4+…+bn)=(a1-6+a3-6+…+an-1-6)+(2a2+2a4+…+2an)=(5+9+…+2n+1-3n)+2×(7+11+…+2n+3)=(5+2n+1)·n22-3n+2×(7+2n+3)·n22=32n2+72n,当n>5时,Tn-Sn=32n2+72n-(n2+4n)=1当n为奇数时,Tn=Tn-1+bn=32(n-1)2+72(n-1)+2n+3-6=32n2+52n-当n>5时,Tn-Sn=32n2+52n-5-(n2+4n)=12n2-32n-5=12(n2-3n-10)=12(即Tn>Sn.综上,当n>5时,Tn>Sn.【自测题】解:(1)证明:∵an+1=2an,n是偶数,an-1,n是奇数,∴a2n=a2n-1-1,a2n+1=2a2n,a2n+2=a2n+1-1,∴a2n-1=a2n+1,又bn=a2n+a2n-1,∴bn=a2n+a2n+1∵bn+1=a2n+2+a2n+1,∴bn+1=a2n+1-1+a2n+1=2a2n+1-1=4a2n-1,∴a2n=bn+1+14,∴即bn+1=2bn-3,∴bn+1-3=2(bn-3).∵b1-3=a1+a2-3=a2=a1-1=2≠0,∴bn-3≠0,∴bn+1-3bn-3=2,∴数列{bn(2)由(1)可知数列{bn-3}是以2为首项,2为公比的等比数列,∴bn-3=2×2n-1=2n,即bn=2n+3,∴a2n-1+a2n=2n+3,∴S2n=2(1-2n)1-2+∴S2n+2=2n+2+3n+1,又a2n=a2n-1-1,∴a2n-1+a2n=2a2n-1-1=2n+3,即a2n-1=2n-1+2,∴a2n+1=2n+2,∴S2n+1=S2n+a2n+1=2n+1+3n-2+2n+2=3·2n+3n.∵S2n+1-S2n=3·2n+3n-(2n+1+3n-2)=2n+2>0,S2n+2-S2n+1=2n+2+3n+1-(3·2n+3n)=2n+1>0,∴当n≥2时,Sn随n的增大而增大,又S1=a1=3,S2=a1+a2=2a1-1=5>S1,∴{Sn}为递增数列.∵S19=3×29+3×9=1563<2023,S20=211+3×10-2=2076>2023,∴满足题意的n的最小值是20.微点2例2解:(1)因为2Sn=3(bn-1),所以当n≥2时,2Sn-1=3(bn-1-1),两式相减得2bn=3bn-3bn-1(n≥2),即bn=3bn-1(n≥2),由2b1=2S1=3(b1-1),解得b1=3,所以数列{bn}是首项为3,公比为3的等比数列,所以bn=3n.在等差数列{cn}中,由c1+c2+c3=27,得3c2=27,解得c2=9,则公差d=c2-c1=4,故cn=c1+(n-1)d=4n+1.所以{bn}和{cn}的通项公式分别为bn=3n,cn=4n+1.(2)设数列{cn}的第m项与数列{bn}的第k项相同,即cm=bk,m,k∈N*,则4m+1=3k=(4-1)k=Ck0×4k+(-1)1Ck1×4k-1+(-1)2Ck2×4k-2+…+(-1)k-1Ckk显然Ck0×4k+(-1)1Ck1×4k-1+(-1)2Ck2×4k-2+…+(-1)k-1Ckk-1×4是4的整数倍,要使4m+1=3k成立,只需k为偶数,因此数列{bn}与{cn}的公共项为b2n=32n=所以T20=9×92×93×…×920=91+2+3+…+20=9210.【自测题】1.162-n+22n[解析]依题意得an=12n,由等差数列的性质得x21+x22=a2+a3,x22-x21=a3-a24-1,即x21+x22=38,x22-x21=-124,解得x22=16.由题得x11=a1+a22=38,x21+x22=a2+a3=38,当n≥3且n为定值时,显然数列{xni}(i∈N*,i≤n)是等差数列,其前n项和记为Tn,则Tn=xn1+xn2+…+xnn=xn1+xnn2·n=an+an+12·n=n212n+12n+1=34·n2n.令T1=x11,T2=x21+x22,则T1=38=34×12,T2=38=34×222,均满足上式,所以Tn=34·n2n,故Sn=x11+x21+x22+…+xn1+xn2+…+xnn=T1+T2+T3+…+T2.解:(1)由Sn+1=Sn+an+4,可得an+1-an=4,故{an}是公差d=4的等差数列,又a2=5,即a1+d=5,所以a1=1,故an=1+(n-1)×4=4n-3.因为{bn}是等比数列,b2=9,b1+b3=30,公比q>1,所以b1q=9,b1+b1q2=30,可得q=3,b1=3,故bn=3n.(2)由(1)知a20=77,令bn<77,得n≤3,b1=3,b2=9,b3=27.因为a3=9与b2=9为公共项,a18=69>b3,所以数列{cn}的前20项是由数列{an}的前18项以及b1,b3构成的,故T20=a1+a2+a3+…+a18+b1+b3=18×1+18×172×4+3+27=660微点3例3解:(1)由题意,当n=1时,a1=A1=3×12+3×1当n≥2时,an=An-An-1=3n2+3n2∵当n=1时,a1=3也满足上式,∴an=3n,n∈N*.由an+1-an=32(bn+1-bn),可得3(n+1)-3n=32(bn+1-bn),整理得bn+1-bn=∵b1=2,∴数列{bn}是以2为首项,以2为公差的等差数列,∴Bn=2n+n(n-1)(2)依题意,由an=Bn及an+1-an=32(bn+1-bn可得bn+1=Bn+1-Bn=32(bn+1-bn化简整理得bn+1=3bn,又b1>0,故数列{bn}是以b1为首项,以3为公比的等比数列,∴Bn=b1(1∵bn+1anan+1∴b2a1a2+b3a2a3+b4a3a4+…+bn+1anan+1=1B1-1B2+1B又对任意n∈N*,b2a1a2+b3a2a3+b4a3a4+…+bn+1anan【自测题】解:(1)设{bn}的公差为d.由题意知Sn=2an+bn(n∈N*),令n=1,得S1=2a1+b1=a1,∴b1=-a1=1.令n=2,得S2=2a2+b2=a1+a2,∵a2+b3=0,∴b2-b3=-1=-d,即d=1,∴bn=n,∴Sn=2an+n,∴Sn-1=2an-1+n-1(n≥2),两式相减得an=2an-2an-1+1(n≥2),∴an=2an-1-1(n≥2),∴an-1=2(an-1-1)(n≥2),又a1-1=-2,∴{an-1}是首项为-2,公比为2的等比数列,∴an-1=-2n,∴an=-2n+1.(2)由c2n=c2n-1+b1,c2n+1=c2n+an可得c2n+1=c2n-1-2n+2,则当n≥2时,c2n-1=c2n-3-2n-1+2,c2n-3=c2n-5-2n-2+2,…,c3=c1-21+2,以上各式累加可得c2n-1+c2n-3+…+c3=c2n-3+c2n-5+…+c1-2n-1-2n-2-…-21+2(n-1)(n≥2),故c2n-1=c1-2(1-2n-1)1-2+2(n-1)=-2n+2n+1(n≥2),又c1=1满足上式,故则T2n=(c1+c3+c5+…+c2n-1)+(c2+c4+c6+…+c2n)=(c1+c3+c5+…+c2n-1)+(c1+1+c3+1+c5+1+…+c2n-1+1)=2(c1+c3+c5+…+c2n-1)+n,又c1+c3+c5+…+c2n-1=-(21+22+…+2n)+(3+5+…+2n+1)=2-2n+1+n2+2n,∴T2n=4-2n+2+2n2+5n.1.解:(1)由题知,a1=1,a2=a1+1=2,a3=a2+2=4,a4=a3+1=5……数列{an}的奇数项构成以1为首项,3为公差的等差数列,所以当n为奇数时,an=1+n+12-1×数列{an}的偶数项构成以2为首项,3为公差的等差数列,所以当n为偶数时,an=2+n2-1×3因为bn=a2n,所以b1=a2=2,b2=a4=5,bn=a2n=3×2n-22=(2)由(1)知{an}的前20项和S20=a1+a2+…+a20=(a1+a3+…+a19)+(a2+a4+…+a20)=10×1+10×92×3+10×2+10×92×3=2.解:(1)因为3a2=3a1+a3,所以3d=a3=a1+2d,即a1=d,故an=nd.所以bn=n2+nnd=n+1d,{bn}是首项为2d,公差为1d的等差数列,所以Sn因为S3+T3=21,所以3×4d2+3×6即2d2-7d+3=0,解得d=3或d=12(舍去故{an}的通项公式为an=3n.(2)若{bn}为等差数列,则2b2=b1+b3,即2·2×3a1+d=即a12-3a1d+2d2=0,所以a1=d或a1=当a1=d时,an=nd,bn=n+1故Sn=n(n+1)d2又S99-T99=99,所以99×100d2-99×102即50d2-d-51=0,解得d=5150或d=-1(舍去)当a1=2d时,an=(n+1)d,bn=nd,故Sn=n(n+3)d又S99-T99=99,所以99×102d2-99×100即51d2-d-50=0,解得d=-5051(舍去)或d=1(舍去)综上,d=5150

高分提能二数列与其他知识的交汇问题例1(1)4(2)①1②88[解析](1)设等比数列{an}的公比为q,q>0,因为an>0,m∈N*,所以由基本不等式得,1=am+2+2am≥2am+2×2am=22q2=22q,所以q≤24,当且仅当am+2=2am,即am+1=2时等号成立.因为116≤a5=a1q4≤244(2)①112+122+132+…+1102<11+12×1+13×2+…+110×9=1+1-12+12-13+…+19-110=1+1-110=1910,112+122+132+…+1102>1+12×3+13×4+…+110×11=1+12-13+13-14+②当n=1时,S1=12a1+1a1=a1,即a12=1,因为a1>0,所以a1=S1=1.当n≥2时,an=Sn-Sn-1,所以2Sn=Sn-Sn-1+1Sn-Sn-1,即Sn2-Sn-12=1,所以{Sn2}是以1为首项,1为公差的等差数列,所以Sn2=n(n∈N*),因为an>0,所以Sn>0,所以Sn=n.当n≥2时,1n+n+1<12n<1n+n-1,即2n+n+1<1n<2n+n-1,所以2(n+1-n)<1n<2(n-n-1).令S=1S1+1S2+…+1S2023,则S>1+2×(3-2+4-3+…+2024-2023)=1+2(2024-2),因为44.98×44.98=2023.2004,44.99×44.99=2024.1001,2<1.42,所以1+2(2024-2)>1+2×(44.98-1.42)=88.12;S<1+2×(2-1+3-2+4-3+…+2023-2022)=1+2(2023-1),因为44.97×44【自测题】解:(1)设{an}的公差为d(d>则a1=a2-d=2-d,a3=a2+d=2+d.∵a1+a3=10,∴(2-d)2+(2+d)2=10,解得d=1或d=-1(舍),则a1=1,∴an=a1+(n-1)d=n(2)证明:∵ai+ai+1=i2+(i+1)2=2i2+2i+1>2i(i+1),∴1ai+∴∑i=1n1ai+例2A[解析]因为f12=33,fn+12=3×fn2(n∈N*),即fn+12fn2=3,所以当n≥2时,fn2=fn2fn-12×fn-12fn-22×…×f22f12×f12=(3)n-1×33=(3)n-2=3n-22,又当n=1时,f12=33,满足上式,故fn2=3n-22,则f(n)=32n-22=3n-1,an=log3f(n)=log33n-1=n-1,所以a12023+a20242023=a22023+a20232023=…=a【自测题】ACD[解析]对于A,因为fnπ2-x=sin2nπ2-x+cos2nπ2-x=cos2nx+sin2nx=fn(x),所以对任意的正整数n,函数fn(x)的图象都关于直线x=π4对称,所以A正确.对于B,当n=1时,f1(x)=1.当n>1时,设sin2x=t,则cos2x=1-t,令h(t)=tn+(1-t)n,可得h'(t)=ntn-1-n(1