2024-2025學(xué)年高二數(shù)學(xué)選擇性必修第一冊（配湘教版）培優(yōu)課 數(shù)列求和

培優(yōu)課數(shù)列求和A級必備知識基礎(chǔ)練1.若數(shù)列{an}的通項公式是an=(-1)n·(3n-2),則它的前100項之和S100=()A.150 B.120 C.-120 D.-1502.已知數(shù)列{an}的前n項和Sn=2n-1,則數(shù)列{an2}的前n項和Tn=(A.(2n-1)2 B.4n-1C.4n-13.數(shù)列{an}的通項公式為an=1n+n+1,若{an}的前n項和為5,則n4.設(shè)等差數(shù)列{an-bn}的公差為2,等比數(shù)列{an+bn}的公比為2,且a1=2,b1=1.求:(1)數(shù)列{an}的通項公式;(2)數(shù)列{2an+2n}的前n項和Sn.5.已知數(shù)列{an}是等差數(shù)列,a1=1,a2+a3+…+a10=144.(1)求數(shù)列{an}的通項公式;(2)若bn=1anan+1,設(shè)Sn是數(shù)列{bn}的前n項和B級關(guān)鍵能力提升練6.數(shù)列{an},{bn}滿足anbn=1,an=n2+5n+6,n∈N+,則{bn}的前10項之和為()A.413 B.513 C.8397.已知在前n項和為Sn的數(shù)列{an}中,a1=1,an+1=-an-2,則S101=()A.-97 B.-98 C.-99 D.-1008.記Sn為數(shù)列{an}的前n項和,若a1=1,a2=2,且an+2-an=1+(-1)n+1,則S100的值為()A.5050 B.2600 C.2550 D.24509.已知等差數(shù)列{an}的前n項和為Sn,a2=3,a7=13.(1)求數(shù)列{an}的通項公式;(2)求證:當(dāng)n∈N+時,Sn2=(Sn)10.已知數(shù)列{an}的前n項和為Sn,且滿足Sn=2an-1,n∈N+.(1)證明數(shù)列{an}是等比數(shù)列,并求{an}的通項公式;(2)設(shè)bn=n·an,求數(shù)列{bn}的前n項和Tn.C級學(xué)科素養(yǎng)創(chuàng)新練11.(多選題)已知數(shù)列{an}是等差數(shù)列,{bn}是等比數(shù)列,a1=1,b1=2,a2+b2=7,a3+b3=13.記cn=an,n為奇數(shù),bn,n為偶數(shù),數(shù)列{cn}A.an=2n-1B.bn=2nC.S9=1409D.S2n=2n2-n+43(4n-12.在等比數(shù)列{an}中,an>0,公比q∈(0,1),且a1a5+2a3a5+a2a8=25,a3與a5的等比中項為2.(1)求數(shù)列{an}的通項公式;(2)設(shè)bn=5-log2an,數(shù)列{bn}的前n項和為Sn,設(shè)Tn=1S1+1S2+…+

培優(yōu)課數(shù)列求和1.AS100=a1+a2+a3+…+a99+a100=-1+4-7+…+(-295)+298=50×3=150.故選A.2.C由等比數(shù)列前n項和的性質(zhì)可知數(shù)列{an}為等比數(shù)列,且an=Sn-Sn-1=2n-1,則an2=4n-1,該數(shù)列{an2}是以1為首項,以4為公比的等比數(shù)列,其前n項和Tn=43.35依題意得an=1n所以Sn=(2-1)+(3-2)+…+(n+1-又因為Sn=n+1-1=5,所以n=354.解(1)因為a1=2,b1=1,所以a1-b1=1,a1+b1=3,依題意可得an-bn=1+2(n-1)=2n-1,an+bn=3×2n-1,故an=2n(2)由(1)可知2an+2n=2n-1+5×2n-1,故Sn=(1+3+…+2n-1)+5×(1+2+…+2n-1)=n(1+2n-1)2+5×(2n-1)=5×5.解(1)因為在等差數(shù)列{an}中,a2+a3+…+a10=144,a1=1,所以9+45d=144,所以d=3.所以數(shù)列{an}的通項公式an=3n-2.(2)因為bn=1anan+1=1(3n-2)(3n+1)=1313n-2-13n+16.D因為anbn=1,an=n2+5n+6,所以bn=1n2+5n+6=1n+2-1n+3,故{bn7.C由an+1=-an-2,得an+an+1=-2,則S101=a1+(a2+a3)+…+(a100+a101)=1-2×50=-99.故選C.8.B當(dāng)n為奇數(shù)時,an+2-an=2,數(shù)列{a2n-1}是首項為1,公差為2的等差數(shù)列;當(dāng)n為偶數(shù)時,an+2-an=0,數(shù)列{a2n}是首項為2,公差為0的等差數(shù)列,即常數(shù)列.則S100=(a1+a3+…+a99)+(a2+a4+…+a100)=50+50×492×2+50×2=2600.9.(1)解設(shè)等差數(shù)列{an}的公差為d,由a2=3,a7=13,可得a1+所以an=1+2(n-1)=2n-1.故數(shù)列{an}的通項公式為an=2n-1.(2)證明由(1)有Sn=n[1+(所以Sn2=(n2)2=n4,(Sn)2=(n2)2=n故當(dāng)n∈N+時,Sn2=(Sn)10.解(1)當(dāng)n=1時,a1=S1=2a1-1,可得a1=1;當(dāng)n>1時,an=Sn-Sn-1=2an-1-(2an-1-1),即an=2an-1.則數(shù)列{an}是首項為1,公比為2的等比數(shù)列,可得an=2n-1.(2)∵bn=n·an=n·2n-1,∴Tn=1×20+2×21+3×22+…+n×2n-1,①∴2Tn=1×2+2×22+3×23+…+(n-1)×2n-1+n×2n,②①-②得-Tn=1+2+22+23+…+2n-1-n×2n=1×(1-2n)1-2-n×2n∴Tn=(n-1)×2n+1.11.ABD設(shè)數(shù)列{an}的公差為d,數(shù)列{bn}的公比為q,依題意有1+d+2q=7,1+2d+2q2=13,得d=2,q=2,故an=2n-1,bn=2n,故A,B正確;則c2n-1=a2n-1=4n-3,c2n=b2n=4n,所以數(shù)列{cn}的前2n項和S2n=(a1+a3+…+a2n-1)+(b2+b4+…+b2n)=n(1+4n-312.解(1)∵a1a5+2a3a5+a2a8=25,∴a32+2a3a5+a5又an>0,∴a3+a5=5.又