新教材2023_2024学年高中数学第一章数列培优课2数列的求和问题分层作业课件北师大版选择性必修第二册

这是一份新教材2023_2024学年高中数学第一章数列培优课2数列的求和问题分层作业课件北师大版选择性必修第二册，共36页。

第1章培优课2　数列的求和问题123456789101112131415161718C1234567891011121314151617182.已知数列{an}的通项公式an=log3 ,设其前n项和为Sn,则使Sn<-4成立的最小正整数n等于(　　)A.83 B.82 C.81 D.80C解析 由题意可得an=log3 =log3n-log3(n+1),故Sn=log31-log32+log32-log33+…+log3n-log3(n+1)=-log3(n+1)<-4,即log3(n+1)>4,解得n>34-1=80,所以使Sn<-4成立的最小正整数n等于81.故选C.123456789101112131415161718A.1 B.2 C.3 D.4 B1234567891011121314151617181234567891011121314151617184.设 ,数列{an}的前n项和Sn=9,则n=　　　　.  99 1234567891011121314151617185.若数列{an}满足a1=1,an+1=2an(n∈N+),则a4=　　　　 ;前8项的和S8=　　　　 .(用数字作答) 8255解析 根据题意,数列{an}满足a1=1,an+1=2an(n∈N+),则数列{an}是首项为1,公比为2的等比数列,则an=a1qn-1=2n-1,故a4=23=8,1234567891011121314151617186.已知在等差数列{an}中,a4=0,a1+a2=-5.(1)求{an}的通项公式;(2)若cn=3nan,求数列{cn}的前n项和Sn.123456789101112131415161718(2)因为cn=(n-4)3n,所以Sn=-3×31+(-2)×32+(-1)×33+…+(n-5)×3n-1+(n-4)×3n,所以3Sn=-3×32+(-2)×33+(-1)×34+…+(n-5)×3n+(n-4)×3n+1,两式相减得-2Sn=-3×31+32+33+34+…+3n-(n-4)×3n+1,所以-2Sn=-12+(3+32+33+34+…+3n)-(n-4)×3n+11234567891011121314151617187.已知数列 是等比数列,a1=1且a2,a3+2,a4成等差数列.(1)求数列{an}的通项公式;(2)设bn= ,求数列{bn}的前n项和Sn.123456789101112131415161718解 (1)设数列 的公比为q,∵a1+1=2,∵2(a3+2)=a2+a4,∴2(2q2+1)=2q-1+2q3-1,∴4q2+2=2q+2q3-2,即4(q2+1)=2q(q2+1),解得q=2.∴an+1=2·2n-1=2n,∴an=2n-1.1234567891011121314151617181234567891011121314151617188.(多选题)[2023江苏南京大学附属中学校考期末]设数列{an}的前n项和为Sn,且Sn=2an-1,bn=log2an+1,则(　　)A.数列{an}是等比数列B.an=(-2)n-1ACD 解析 由已知Sn=2an-1,当n=1时,可得a1=1,选项A,当n≥2时,Sn-Sn-1=an=2an-2an-1,an=2an-1,可得数列{an}是1为首项,2为公比的等比数列,故A正确;选项B,由选项A可得an=2n-1,故B错误;故选ACD. 1234567891011121314151617181234567891011121314151617189.已知数列{3n+1}与数列{4n-1},其中n∈N+.它们的公共项由小到大组成新的数列{an},则{an}前25项的和为(　　)A.3 197 B.3 480 C.3 586 D.3 775D 解析 数列{3n+1}(n∈N+)的各项为:4,7,10,13,16,19,22,25,28,31,…,数列{4n-1}(n∈N+)的各项为:3,7,11,15,19,23,27,31,35,39,…,由题意可知,数列{an}的各项为:7,19,31,…,所以数列{an}为等差数列,且首项为7,公差为19-7=12,因此,数列{an}的前25项的和为7×25+ =3 775.故选D.12345678910111213141516171810.已知数列{an}的前n项和为Sn,a1=2,am+n=aman,则S6=(　　)A.12 B.27-1 C.27 D.27-2D 12345678910111213141516171811.(多选题)[2023安徽蚌埠二中阶段练习]设等差数列{an}的前n项和为Sn,且 恒成立,则λ的值不可能是(　　)A.1 B.0 C.-1 D.2BC123456789101112131415161718123456789101112131415161718123456789101112131415161718AD12345678910111213141516171812345678910111213141516171812345678910111213141516171813.已知数列{an}的前n项和为Sn,且满足Sn=2an-1(n∈N+),则数列{nan}的前n项和Tn为　　　　　　. (n-1)2n+1 123456789101112131415161718解析 ∵Sn=2an-1(n∈N+),∴当n=1时,a1=2a1-1,解得a1=1,当n≥2时,an=Sn-Sn-1=2an-1-(2an-1-1),化为an=2an-1,∴数列{an}是首项为1,公比为2的等比数列,∴an=2n-1.∴nan=n·2n-1.则数列{nan}的前n项和Tn=1+2×2+3×22+…+n·2n-1.∴2Tn=2+2×22+…+(n-1)·2n-1+n·2n,∴-Tn=1+2+22+…+2n-1-n·2n= -n·2n=(1-n)·2n-1,∴Tn=(n-1)2n+1.12345678910111213141516171814.设等差数列{an}满足a2=5,a6+a8=30,则an=　　　　　,数列 的前n项和为　　　　. 2n+1 解析 设数列{an}的公差为d.∵a6+a8=30=2a7,∴a7=15,∵a7-a2=5d,a2=5,∴d=2,∴an=a2+(n-2)d=2n+1.12345678910111213141516171815.已知{an}是等比数列,{bn}是等差数列,且a1=1,b1=3,a2+b2=7,a3+b3=11.(1)求数列{an}和{bn}的通项公式;(2)设cn= ,n∈N+,求数列{cn}的前n项和Tn.12345678910111213141516171812345678910111213141516171812345678910111213141516171816.已知函数f(x)= (x∈R),正项等比数列{an}满足a50=1,则f(ln a1)+f(ln a2)+…+f(ln a99)的值是多少?123456789101112131415161718因为数列{an}是等比数列,所以a1a99=a2a98=…=a49a51= =1,即ln a1+ln a99=ln a2+ln a98=…=ln a49+ln a51=2ln a50=0.所以f(ln a1)+f(ln a99)=f(ln a2)+f(ln a98)=…=f(ln a49)+f(ln a51)=2f(ln a50)=1.设S99=f(ln a1)+f(ln a2)+f(ln a3)+…+f(ln a99),①又S99=f(ln a99)+f(ln a98)+f(ln a97)+…+f(ln a1),②①+②,得2S99=99,所以S99= .12345678910111213141516171817.已知首项为-2的等差数列{an}的前n项和为Sn,数列{bn}满足 Sn=2n(log2bn-2)(n∈N+),b3=8.(1)求an与bn;123456789101112131415161718(1)解 设等差数列{an}的公差为d,由题意知S3=2×3×(log2b3-2)=2×3×(3-2)=6,因为S3=a1+a2+a3=3a2=6,解得a2=2,所以d=a2-a1=2-(-2)=4,因为Sn=2n(log2bn-2)(n∈N+),所以2n2-4n=2n(log2bn-2),解得bn=2n.12345678910111213141516171812345678910111213141516171818.已知正项数列{an},其前n项和为Sn,an=1-2Sn(n∈N+).(1)求数列{an}的通项公式;123456789101112131415161718123456789101112131415161718