2025高考数学二轮复习-专题突破练12 数列求和【课件】

专题突破练12数列求和20251234567891011一、选择题1.(2024·浙江杭州二模)设数列{an},{bn}满足a1=b1=1,an+bn+1=2n,an+1+bn=2n.设Sn为数列{an+bn}的前n项的和,则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.故选A.1234567891011A.511 B.677

C.1021 D.2037B1234567891011C1234567891011解析

设等差数列{an}的公差为d(d≠0),因为a3+a8=20,且a5是a2与a14的等比中项,12345678910114.(2024·安徽三模)记数列{an}的前n项和为Sn,若a1=1,Sn+Sn+1=3n2+2n+1,则S20=(

)A.590 B.602

C.630

D.650A1234567891011解析

因为Sn+Sn+1=3n2+2n+1,所以Sn+1+Sn+2=3(n+1)2+2(n+1)+1,两式相减可得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,满足上式,故an+an+1=6n-1,所以S20=(a1+a2)+(a3+a4)+…+(a19+a20)=5+17+29+…+113==590.故选A.12345678910115.(2024·江西模拟)在学习完“错位相减法”后,善于观察的同学发现对于“等差×等比数列”此类数列求和,也可以使用“裂项相消法”求解.例如an=(n+1)·2n=(-n+1)·2n-(-n)·2n+1,故数列{an}的前n项和Sn=a1+a2+a3+…+an=0×21-(-1)×22+(-1)×22-(-2)×23+…+(-n+1)·2n-(-n)·2n+1=n·2n+1.记数列

的前n项和为Tn,利用上述方法求T20-6=(

)B12345678910111234567891011A.18 B.12 C.9 D.6C1234567891011两式相减得an+1=an+2·an+1-an+1·an,两边同时除以an+1(an+1≠0)得an+2-an=1,由上式可知数列{an}的奇数项和偶数项分别成等差数列,公差为1,=(1+2+3+4+5)-(a1+a1+1+a1+2+a1+3+a1+4+a1+5)=-6a1,由此数列的奇数项公式为a2n-1=a1+(n-1),又由an≠0,所以可以判断a1一定不能为负整数,即只能有-6a1=9.故选C.1234567891011二、选择题7.(2024·安徽淮北二模)已知数列{an},{bn}的前n项和分别为Sn,Tn,若an=2n-1,Tn=2n+1-2,则下列说法正确的是(

)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}是首项为b1=2,公比q=2的等比数列,∴b10=210=1

024.故选项B正确.12345678910111234567891011BC1234567891011解析

根据题意,列举可得,数列{an}的前若干项分别为1,2,3,3,4,5,6,6,….不难发现,a4k=3k,a4k-1=3k,a4k-2=3k-1,a4k-3=3k-2.对于A,根据数列列举可得1,2,3,3,4,5,6,6,7,8,9,9,10,11,12,12,13,14,15,15,16,17,18,18,….即落在区间[23,24)上的有8,9,9,10,11,12,12,13,14,15,15共有11项,因此b3=11.故A错误.1234567891011对于D,根据数列列举可得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、填空题9.(2024·江西宜春模拟)已知数列{an}是等差数列,

记Sn,Tn分别为{an},{bn}的前n项和,若S3=18,T2=10,则T20=

.

3701234567891011解析

设等差数列{an}的公差为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、解答题10.(15分)(2024·浙江宁波二模)已知等差数列{an}的公差为2,记数列{bn}的前n项和为Sn,b1=0,b2=2且满足bn+1=2Sn+an.(1)证明:数列{bn+1}是等比数列;(2)求数列{anbn}的前n项和Tn.1234567891011(1)证明

当n≥2时,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,所以当n≥1时,bn+1=3bn+2,即bn+1+1=3(bn+1).又b1+1=1≠0,所以bn+1≠0,所以数列{bn+1}成等比数列.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,则3M=1×31+2×32+3×33+…+n×3n,123456789101111.(15分)(2024·山东潍坊三模)已知正项等差数列{an}的公差为2,前n项和为Sn,且S1+1,S2,S3+1成等比数列.(1)求数列{an}的通项公