2022新高考總復(fù)習(xí)《數(shù)學(xué)》(人教)第六章　數(shù)列課時(shí)規(guī)范練33　數(shù)列求和

1、 課時(shí)規(guī)范練33數(shù)列求和基礎(chǔ)鞏固組1.(2020山東濱州模擬)若數(shù)列an的通項(xiàng)公式為an=2n+2n-1,則該數(shù)列的前10項(xiàng)和為() A.2 146B.1 122C.2 148D.1 1242.已知函數(shù)f(n)=n2,n為奇數(shù),-n2,n為偶數(shù),且an=f(n)+f(n+1),則a1+a2+a3+a100等于()A.0B.100C.-100D.10 2003.在數(shù)列an中,若an+1+(-1)nan=2n-1,則數(shù)列an的前12項(xiàng)和等于()A.76B.78C.80D.824.已知數(shù)列an,若an+1=an+an+2(nN*),則稱數(shù)列an為“凸數(shù)列”.已知數(shù)列bn為“凸數(shù)列”,且b1=1,b2

2、=-2,則數(shù)列bn的前2 020項(xiàng)和為()A.5B.-5C.0D.-45.(多選)公差為d的等差數(shù)列an滿足a2=5,a6+a8=30,則下面結(jié)論正確的是()A.d=2B.an=2n+1C.1an2-1=141n+1n+1D.1an2-1的前n項(xiàng)和為n4(n+1)6.(多選)數(shù)列an滿足a1=1,且對(duì)任意的nN*都有an+1=an+n+1,則()A.an=n(n+1)2B.數(shù)列1an的前100項(xiàng)和為200101C.數(shù)列1an的前100項(xiàng)和為99100D.數(shù)列an的第100項(xiàng)為50 0507.(2020山東德州調(diào)研)已知Tn為數(shù)列2n+12n的前n項(xiàng)和,若mT10+1 013恒成立,則整數(shù)m的最

3、小值為()A.1 026B.1 025C.1 024D.1 0238.(2020河北“五個(gè)一”名校質(zhì)檢)若f(x)+f(1-x)=4,an=f(0)+f1n+fn-1n+f(1)(nN*),則數(shù)列an的通項(xiàng)公式為.9.(2020安徽阜陽(yáng)太和模擬)設(shè)Sn是數(shù)列an的前n項(xiàng)和,且a1=1,an+1+SnSn+1=0,則Sn=,數(shù)列SnSn+1的前n項(xiàng)和Tn為.10.(2020山東濰坊高三上期末)已知各項(xiàng)均不相等的等差數(shù)列an的前4項(xiàng)和為10,且a1,a2,a4是等比數(shù)列bn的前3項(xiàng).(1)求an,bn;(2)設(shè)cn=bn+1an(an+1),求cn的前n項(xiàng)和Sn.11.(2020山東棗莊滕州高三上

4、期末)已知等比數(shù)列an滿足a1,a2,a3-a1成等差數(shù)列,且a1a3=a4.等差數(shù)列bn的前n項(xiàng)和Sn=(n+1)log2an2.(1)求an,bn;(2)求數(shù)列anbn的前n項(xiàng)和Tn.綜合提升組12.(2020河南鄭州模擬)數(shù)列an滿足a1=1,且對(duì)任意的m,nN*,都有am+n=am+an+mn,則1a1+1a2+1a3+1a2 018=()A.2 0172 018B.2 0182 019C.4 0342 018D.4 0362 01913.(2020廣東肇慶模擬)已知各項(xiàng)均為正數(shù)的數(shù)列an滿足a1=1,a2=12,1an+1=1an1an+2(nN*),那么a1a3+a2a4+a3a5

5、+anan+2=.14.(2020山東九校高三上學(xué)期聯(lián)考)已知在數(shù)列an中,a1=12,其前n項(xiàng)和Sn滿足Sn2-anSn+an=0(n2),則a2=,S2 019=.15.(2020新高考全國(guó)1,18)已知公比大于1的等比數(shù)列an滿足a2+a4=20,a3=8.(1)求an的通項(xiàng)公式;(2)記bm為an在區(qū)間(0,m(mN*)中的項(xiàng)的個(gè)數(shù),求數(shù)列bm的前100項(xiàng)和S100.創(chuàng)新應(yīng)用組16.(多選)已知函數(shù)f(x)=12(x2+a)的圖象在點(diǎn)Pn(n,f(n)(nN*)處的切線ln的斜率為kn,直線ln交x軸,y軸分別于點(diǎn)An(xn,0),Bn(0,yn),且y1=-1.以下結(jié)論中,正確的結(jié)論

6、是()A.a=-1B.記函數(shù)g(n)=xn(nN*),則函數(shù)g(n)先減后增,且最小值為1C.當(dāng)nN*時(shí),yn+kn+12ln(1+kn)D.當(dāng)nN*時(shí),記數(shù)列1|yn|kn的前n項(xiàng)和為Sn,則SnSk+1且Sk+1T10+1 013恒成立,整數(shù)m的最小值為1 024.8.an=2(n+1)由f(x)+f(1-x)=4,可得f(0)+f(1)=4,f1n+fn-1n=4,所以2an=f(0)+f(1)+f1n+fn-1n+fn-1n+f1n+f(1)+f(0)=4(n+1),即an=2(n+1).9.1nnn+1an+1=Sn+1-Sn,an+1+SnSn+1=0,Sn+1-Sn+SnSn+1

7、=0,1Sn+1-1Sn=1.又1S1=1a1=1,1Sn是以1為首項(xiàng),1為公差的等差數(shù)列,1Sn=n,Sn=1n.SnSn+1=1n(n+1)=1n-1n+1,Tn=1-12+12-13+1n-1n+1=1-1n+1=nn+1.10.解 (1)設(shè)數(shù)列an的公差為d,由題意知a1+a2+a3+a4=4a1+4(4-1)2d=4a1+6d=10.又因?yàn)閍1,a2,a4成等比數(shù)列,所以a22=a1a4,即(a1+d)2=a1(a1+3d),化簡(jiǎn)得d2=a1d,又因?yàn)閐0,所以a1=d.由得a1=1,d=1,所以an=n.b1=a1=1,b2=a2=2,q=b2b1=2,所以bn=2n-1.(2)由

8、(1)及cn=bn+1an(an+1)可得,cn=2n-1+1n(n+1)=2n-1+1n-1n+1,所以Sn=20+21+2n-1+1-12+12-13+1n-1n+1=1-2n1-2+1-1n+1=2n-1n+1,所以數(shù)列cn的前n項(xiàng)和Sn=2n-1n+1.11.解 (1)設(shè)an的公比為q,bn的公差為d.因?yàn)閍1,a2,a3-a1成等差數(shù)列,所以2a2=a1+(a3-a1),即2a2=a3.因?yàn)閍20,所以q=a3a2=2.因?yàn)閍1a3=a4,所以a1=a4a3=q=2.因此an=a1qn-1=2n.由題意,Sn=(n+1)log2an2=(n+1)n2.所以b1=S1=1,b1+b2=

9、S2=3,從而b2=2.所以bn的公差d=b2-b1=2-1=1.所以bn=b1+(n-1)d=1+(n-1)1=n.(2)令cn=anbn,則cn=n2n.因此Tn=c1+c2+cn-1+cn=121+222+323+(n-1)2n-1+n2n.又因?yàn)?Tn=122+223+324+(n-1)2n+n2n+1,兩式相減得-Tn=2+22+23+2n-n2n+1=2-2n21-2-n2n+1=2n+1-2-n2n+1=(1-n)2n+1-2.所以Tn=(n-1)2n+1+2.12.D因?yàn)閍1=1,且對(duì)任意的m,nN*都有am+n=am+an+mn,令m=1,則有an+1=an+n+1,即an+

10、1-an=n+1,用累加法可得an=a1+(n-1)(n+2)2=n(n+1)2,所以1an=2n(n+1)=21n-1n+1,所以1a1+1a2+1a3+1a2 018=21-12+12-13+12 018-12 019=21-12 019=4 0362 019.13.131-14n由1an+1=1an1an+2(nN*),可得an+12=anan+2,數(shù)列an為等比數(shù)列.a1=1,a2=12,q=12,an=12n-1,anan+2=12n-112n+1=14n,a1a3=14,a1a3+a2a4+a3a5+anan+2=14+142+14n=14(1-14n)1-14=131-14n.1

11、4.-1612 020由題意,知Sn2-anSn+an=0(n2),令n=2,則S22-a2S2+a2=0,即(a2+12)2-a2a2+12+a2=0,化簡(jiǎn)得32a2+14=0,所以a2=-16.因?yàn)镾n2-anSn+an=0(n2),an=Sn-Sn-1(n2),所以SnSn-1+Sn-Sn-1=0(n2),整理得1Sn-1Sn-1=1(n2),又因?yàn)?S1=1a1=2,所以1Sn是一個(gè)以2為首項(xiàng),1為公差的等差數(shù)列,所以1Sn=n+1,所以Sn=1n+1,所以S2 019=12 020.15.解 (1)設(shè)an的公比為q.由題設(shè)得a1q+a1q3=20,a1q2=8.解得q=12(舍去),

12、q=2.因?yàn)閍1q2=8,所以a1=2.所以an的通項(xiàng)公式為an=2n.(2)由題設(shè)及(1)知b1=0,且當(dāng)2nm1時(shí),xn0,函數(shù)g(n)為增函數(shù),當(dāng)n=1時(shí),函數(shù)取最小值,且最小值為1.函數(shù)g(n)是單調(diào)遞增的,且最小值為1,故B不正確.在ln中,令x=0,得yn=-n2+12(n2-1)=-12(n2+1), yn+kn+12=-12n2+n,當(dāng)n=1時(shí),y1+k1+12=12=lneln 1=0,故C正確.1|yn|kn=2n2+1n2n2,Sn2112+122+132+1n2.當(dāng)n=1時(shí),S1=11時(shí),1n21n(n-1)=1n-1-1n,SnSk+1,即SkSk+ak+1,則ak+10;同理,若使Sk+1Sk+2,即Sk+10.若選b1+b3=a2,則a2=-10,a5=-1,d=3,a1=-13,ak=3k-16,a