2025高考數(shù)學(xué)二輪復(fù)習(xí)-專(zhuān)題突破練12 數(shù)列求和【課件】

專(zhuān)題突破練12數(shù)列求和20251234567891011一、選擇題1.(2024·浙江杭州二模)設(shè)數(shù)列{an},{bn}滿足a1=b1=1,an+bn+1=2n,an+1+bn=2n.設(shè)Sn為數(shù)列{an+bn}的前n項(xiàng)的和,則S7=(

)A.110 B.120 C.288

D.306A解析

S7=a1+b1+a2+b2+a3+b3+a4+b4+a5+b5+a6+b6+a7+b7=a1+b1+(a2+b3)+(b2+a3)+(a4+b5)+(b4+a5)+(a6+b7)+(b6+a7)=1+1+2×2+22+2×4+24+2×6+26=2+4+4+8+16+12+64=110.故選A.1234567891011A.511 B.677

C.1021 D.2037B1234567891011C1234567891011解析

設(shè)等差數(shù)列{an}的公差為d(d≠0),因?yàn)閍3+a8=20,且a5是a2與a14的等比中項(xiàng),12345678910114.(2024·安徽三模)記數(shù)列{an}的前n項(xiàng)和為Sn,若a1=1,Sn+Sn+1=3n2+2n+1,則S20=(

)A.590 B.602

C.630

D.650A1234567891011解析

因?yàn)镾n+Sn+1=3n2+2n+1,所以Sn+1+Sn+2=3(n+1)2+2(n+1)+1,兩式相減可得an+1+an+2=6n+5=6(n+1)-1.由a1=1,S1+S2=3×12+2×1+1=6,解得a2=4,所以a1+a2=5,滿足上式,故an+an+1=6n-1,所以S20=(a1+a2)+(a3+a4)+…+(a19+a20)=5+17+29+…+113==590.故選A.12345678910115.(2024·江西模擬)在學(xué)習(xí)完“錯(cuò)位相減法”后,善于觀察的同學(xué)發(fā)現(xiàn)對(duì)于“等差×等比數(shù)列”此類(lèi)數(shù)列求和,也可以使用“裂項(xiàng)相消法”求解.例如an=(n+1)·2n=(-n+1)·2n-(-n)·2n+1,故數(shù)列{an}的前n項(xiàng)和Sn=a1+a2+a3+…+an=0×21-(-1)×22+(-1)×22-(-2)×23+…+(-n+1)·2n-(-n)·2n+1=n·2n+1.記數(shù)列

的前n項(xiàng)和為T(mén)n,利用上述方法求T20-6=(

)B12345678910111234567891011A.18 B.12 C.9 D.6C1234567891011兩式相減得an+1=an+2·an+1-an+1·an,兩邊同時(shí)除以an+1(an+1≠0)得an+2-an=1,由上式可知數(shù)列{an}的奇數(shù)項(xiàng)和偶數(shù)項(xiàng)分別成等差數(shù)列,公差為1,=(1+2+3+4+5)-(a1+a1+1+a1+2+a1+3+a1+4+a1+5)=-6a1,由此數(shù)列的奇數(shù)項(xiàng)公式為a2n-1=a1+(n-1),又由an≠0,所以可以判斷a1一定不能為負(fù)整數(shù),即只能有-6a1=9.故選C.1234567891011二、選擇題7.(2024·安徽淮北二模)已知數(shù)列{an},{bn}的前n項(xiàng)和分別為Sn,Tn,若an=2n-1,Tn=2n+1-2,則下列說(shuō)法正確的是(

)A.S10=100B.b10=1024ABD1234567891011∵Tn=2n+1-2,∴bn=Tn-Tn-1=2n+1-2-2n+2=2n(n>1,n∈N*),又b1=T1=21+1-2=2,b1=21=2,∴bn=2n(n∈N*),∴{bn}是首項(xiàng)為b1=2,公比q=2的等比數(shù)列,∴b10=210=1

024.故選項(xiàng)B正確.12345678910111234567891011BC1234567891011解析

根據(jù)題意,列舉可得,數(shù)列{an}的前若干項(xiàng)分別為1,2,3,3,4,5,6,6,….不難發(fā)現(xiàn),a4k=3k,a4k-1=3k,a4k-2=3k-1,a4k-3=3k-2.對(duì)于A,根據(jù)數(shù)列列舉可得1,2,3,3,4,5,6,6,7,8,9,9,10,11,12,12,13,14,15,15,16,17,18,18,….即落在區(qū)間[23,24)上的有8,9,9,10,11,12,12,13,14,15,15共有11項(xiàng),因此b3=11.故A錯(cuò)誤.1234567891011對(duì)于D,根據(jù)數(shù)列列舉可得1,2,3,3,4,5,6,6,7,8,9,9,10,11,12,12,13,14,15,15,16,17,18,18,….可得b1=3,b2=5,1234567891011三、填空題9.(2024·江西宜春模擬)已知數(shù)列{an}是等差數(shù)列,

記Sn,Tn分別為{an},{bn}的前n項(xiàng)和,若S3=18,T2=10,則T20=

.

3701234567891011解析

設(shè)等差數(shù)列{an}的公差為d.由S3=18,得a1+a2+a3=a1+a1+d+a1+2d=3a1+3d=18,①由T2=10,得a1-8+2a2+1=a1-8+2(a1+d)+1=10,所以3a1+2d=17,②解得a1=5,d=1,所以an=a1+(n-1)d=5+(n-1)×1=n+4.所以T20=(b1+b3+…+b19)+(b2+b4+…+b20)=(-3-1+1+…+15)1234567891011四、解答題10.(15分)(2024·浙江寧波二模)已知等差數(shù)列{an}的公差為2,記數(shù)列{bn}的前n項(xiàng)和為Sn,b1=0,b2=2且滿足bn+1=2Sn+an.(1)證明:數(shù)列{bn+1}是等比數(shù)列;(2)求數(shù)列{anbn}的前n項(xiàng)和Tn.1234567891011(1)證明

當(dāng)n≥2時(shí),bn+1-bn=2(Sn-Sn-1)+an-an-1=2bn+2,即bn+1=3bn+2.又b1=0,b2=2,也符合b2=3b1+2,所以當(dāng)n≥1時(shí),bn+1=3bn+2,即bn+1+1=3(bn+1).又b1+1=1≠0,所以bn+1≠0,所以數(shù)列{bn+1}成等比數(shù)列.1234567891011(2)解

由(1)易得bn=3n-1-1.由b2=2b1+a1可得a1=2,所以an=2n.故anbn=2n(3n-1-1)=2n×3n-1-2n,所以Tn=2(1×30+2×31+3×32+…+n×3n-1)-n(n+1).令M=1×30+2×31+3×32+…+n×3n-1,則3M=1×31+2×32+3×33+…+n×3n,123456789101111.(15分)(2024·山東濰坊三模)已知正項(xiàng)等差數(shù)列{an}的公差為2,前n項(xiàng)和為Sn,且S1+1,S2,S3+1成等比數(shù)列.(1)求數(shù)列{an}的通項(xiàng)公