2024-2025学年高二数学选择性必修第一册（配湘教版）培优课 数列求和

培优课数列求和A级必备知识基础练1.若数列{an}的通项公式是an=(-1)n·(3n-2),则它的前100项之和S100=()A.150 B.120 C.-120 D.-1502.已知数列{an}的前n项和Sn=2n-1,则数列{an2}的前n项和Tn=(A.(2n-1)2 B.4n-1C.4n-13.数列{an}的通项公式为an=1n+n+1,若{an}的前n项和为5,则n4.设等差数列{an-bn}的公差为2,等比数列{an+bn}的公比为2,且a1=2,b1=1.求:(1)数列{an}的通项公式;(2)数列{2an+2n}的前n项和Sn.5.已知数列{an}是等差数列,a1=1,a2+a3+…+a10=144.(1)求数列{an}的通项公式;(2)若bn=1anan+1,设Sn是数列{bn}的前n项和B级关键能力提升练6.数列{an},{bn}满足anbn=1,an=n2+5n+6,n∈N+,则{bn}的前10项之和为()A.413 B.513 C.8397.已知在前n项和为Sn的数列{an}中,a1=1,an+1=-an-2,则S101=()A.-97 B.-98 C.-99 D.-1008.记Sn为数列{an}的前n项和,若a1=1,a2=2,且an+2-an=1+(-1)n+1,则S100的值为()A.5050 B.2600 C.2550 D.24509.已知等差数列{an}的前n项和为Sn,a2=3,a7=13.(1)求数列{an}的通项公式;(2)求证:当n∈N+时,Sn2=(Sn)10.已知数列{an}的前n项和为Sn,且满足Sn=2an-1,n∈N+.(1)证明数列{an}是等比数列,并求{an}的通项公式;(2)设bn=n·an,求数列{bn}的前n项和Tn.C级学科素养创新练11.(多选题)已知数列{an}是等差数列,{bn}是等比数列,a1=1,b1=2,a2+b2=7,a3+b3=13.记cn=an,n为奇数,bn,n为偶数,数列{cn}A.an=2n-1B.bn=2nC.S9=1409D.S2n=2n2-n+43(4n-12.在等比数列{an}中,an>0,公比q∈(0,1),且a1a5+2a3a5+a2a8=25,a3与a5的等比中项为2.(1)求数列{an}的通项公式;(2)设bn=5-log2an,数列{bn}的前n项和为Sn,设Tn=1S1+1S2+…+

培优课数列求和1.AS100=a1+a2+a3+…+a99+a100=-1+4-7+…+(-295)+298=50×3=150.故选A.2.C由等比数列前n项和的性质可知数列{an}为等比数列,且an=Sn-Sn-1=2n-1,则an2=4n-1,该数列{an2}是以1为首项,以4为公比的等比数列,其前n项和Tn=43.35依题意得an=1n所以Sn=(2-1)+(3-2)+…+(n+1-又因为Sn=n+1-1=5,所以n=354.解(1)因为a1=2,b1=1,所以a1-b1=1,a1+b1=3,依题意可得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)因为在等差数列{an}中,a2+a3+…+a10=144,a1=1,所以9+45d=144,所以d=3.所以数列{an}的通项公式an=3n-2.(2)因为bn=1anan+1=1(3n-2)(3n+1)=1313n-2-13n+16.D因为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,则S101=a1+(a2+a3)+…+(a100+a101)=1-2×50=-99.故选C.8.B当n为奇数时,an+2-an=2,数列{a2n-1}是首项为1,公差为2的等差数列;当n为偶数时,an+2-an=0,数列{a2n}是首项为2,公差为0的等差数列,即常数列.则S100=(a1+a3+…+a99)+(a2+a4+…+a100)=50+50×492×2+50×2=2600.9.(1)解设等差数列{an}的公差为d,由a2=3,a7=13,可得a1+所以an=1+2(n-1)=2n-1.故数列{an}的通项公式为an=2n-1.(2)证明由(1)有Sn=n[1+(所以Sn2=(n2)2=n4,(Sn)2=(n2)2=n故当n∈N+时,Sn2=(Sn)10.解(1)当n=1时,a1=S1=2a1-1,可得a1=1;当n>1时,an=Sn-Sn-1=2an-1-(2an-1-1),即an=2an-1.则数列{an}是首项为1,公比为2的等比数列,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设数列{an}的公差为d,数列{bn}的公比为q,依题意有1+d+2q=7,1+2d+2q2=13,得d=2,q=2,故an=2n-1,bn=2n,故A,B正确;则c2n-1=a2n-1=4n-3,c2n=b2n=4n,所以数列{cn}的前2n项和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