2021版高考數(shù)學(xué)一輪復(fù)習(xí)核心素養(yǎng)測(cè)評(píng)三十九8.4數(shù)列的求和文含解析北師大版

1、核心素養(yǎng)測(cè)評(píng)三十九數(shù)列的求和(30分鐘60分)一、選擇題(每小題5分,共25分)1.數(shù)列an的通項(xiàng)公式是an=1n+n+1,若前n項(xiàng)和為10,則項(xiàng)數(shù)n為()a.120b.99c.11d.121【解析】選a.an=1n+n+1=n+1-n(n+1+n)(n+1-n)=n+1-n,所以a1+a2+an=(2-1)+(3-2)+(n+1-n)=n+1-1=10.即n+1=11,所以n+1=121,n=120.2.已知數(shù)列an的通項(xiàng)公式是an=2n-315n,則其前20項(xiàng)和為()a.380-351-1519b.400-251-1520c.420-341-1520d.440-451-1520【解析】選c

2、.令數(shù)列an的前n項(xiàng)和為sn,則s20=a1+a2+a20=2(1+2+20)-315+152+1520=220(20+1)2-3151-15201-15=420-341-1520.3.已知等比數(shù)列an,a1=1,a4=18,且a1a2+a2a3+anan+1k,則k的取值范圍是()a.12,23b.12,+c.12,23d.23,+【解析】選d.設(shè)等比數(shù)列an的公比為q,則q3=a4a1=18,解得q=12,所以an=12n-1,所以anan+1=12n-112n=122n-1,所以數(shù)列anan+1是首項(xiàng)為12,公比為14的等比數(shù)列,所以a1a2+a2a3+anan+1=121-14n1-1

3、4=231-14n0nn*.世紀(jì)金榜導(dǎo)學(xué)號(hào)(1)求數(shù)列an的通項(xiàng)公式.(2)若bn=3n2n-1nannn*,求數(shù)列bn的前n項(xiàng)和tn.【解析】(1)當(dāng)n=1時(shí),2s1=a1-1a1+2=2a1,因?yàn)閍10,所以a1=2,當(dāng)n2時(shí),2an=2sn-sn-1=an-1an+2-an-1-1an-1+2,所以an+an-1an-an-1-1=0,因?yàn)閍n0,所以an-an-1-1=0,所以an-an-1=1,所以an是以a1=2為首項(xiàng),d=1為公差的等差數(shù)列,所以an=n+1nn*.(2)由(1)得an=n+1,所以bn=3n2n-1nn+1=3n+1n+1-3nn,所以tn=b1+b2+bn-1

4、+bn=322-3+333-322+3nn-3n-1n-1+3n+1n+1-3nn=3n+1n+1-3.10.已知數(shù)列an的各項(xiàng)均為正數(shù),且an2-2nan-(2n+1)=0,nn*.世紀(jì)金榜導(dǎo)學(xué)號(hào)(1)求數(shù)列an的通項(xiàng)公式.(2)若bn=2nan,求數(shù)列bn的前n項(xiàng)和tn.【解析】(1)由an2-2nan-(2n+1)=0得an-(2n+1)(an+1)=0,所以an=2n+1或an=-1,又?jǐn)?shù)列an的各項(xiàng)均為正數(shù),負(fù)值應(yīng)舍去,所以an=2n+1,nn*.(2)因?yàn)閎n=2nan=2n(2n+1),所以tn=23+225+237+2n(2n+1),2tn=223+235+2n(2n-1)+2

5、n+1(2n+1),由-得-tn=6+2(22+23+2n)-2n+1(2n+1)=6+222(1-2n-1)1-2-2n+1(2n+1)=-2+2n+1(1-2n).所以tn=(2n-1)2n+1+2.(15分鐘35分)1.(5分)若數(shù)列an的通項(xiàng)公式是an=(-1)n(3n-2),則a1+a2+a3+a10=()a.15b.12c.-12d.-15【解析】選a.因?yàn)閍n=(-1)n(3n-2),所以a1+a2+a10=-1+4-7+10-13+16-19+22-25+28=(-1+4)+(-7+10)+(-13+16)+(-19+22)+(-25+28)=35=15.【變式備選】已知數(shù)列a

6、n的前n項(xiàng)和為sn,通項(xiàng)公式an=n(-1)n+1,則s17=()a.10b.9c.8d.7【解析】選b.s17=1-2+3-4+5-6+15-16+17=1+(-2+3)+(-4+5)+(-14+15)+(-16+17)=1+1+1+1=9.【一題多解】解決本題還可以采用以下方法:選b.s17=1-2+3-4+5-6+15-16+17=(1+3+17)-(2+4+16)=81-72=9.2.(5分)已知等比數(shù)列an的首項(xiàng)為32,公比為-12,其前n項(xiàng)和為sn,則sn的最大值為()a.34b.23c.43d.32【解析】選d.因?yàn)榈缺葦?shù)列an的首項(xiàng)為32,公比為-12,所以sn=321-12n

7、1-12=1-12n,當(dāng)n取偶數(shù)時(shí),sn=1-12n0,解得q=2,所以bn=2n.設(shè)等差數(shù)列an的公差為d,由b3=a4-2a1可得3d-a1=8,由s11=11b4可得a1+5d=16,聯(lián)立解得a1=1,d=3,由此可得an=3n-2.所以an的通項(xiàng)公式為an=3n-2,bn的通項(xiàng)公式為bn=2n.(2)設(shè)數(shù)列a2nbn的前n項(xiàng)和為tn,由a2n=6n-2得tn=42+1022+1623+(6n-2)2n,2tn=422+1023+1624+(6n-8)2n+(6n-2)2n+1,上述兩式相減得:-tn=42+622+623+62n-(6n-2)2n+1=12(1-2n)1-2-4-(6n

8、-2)2n+1=-(3n-4)2n+2-16.所以tn=(3n-4)2n+2+16.所以數(shù)列a2nbn的前n項(xiàng)和為(3n-4)2n+2+16.1.已知數(shù)列an中,a1=2,點(diǎn)an,an+1在函數(shù)fx=x2+2x的圖像上,其中n=1,2,3,.若bn=1an+1an+2,數(shù)列bn的前n項(xiàng)和為sn,則s2 020+2a2 021=世紀(jì)金榜導(dǎo)學(xué)號(hào)()a.2 020b.20c.2d.1【解析】選d.因?yàn)辄c(diǎn)an,an+1在函數(shù)fx=x2+2x的圖像上,所以an+1=an2+2an,所以1an+1=121an-1an+2,所以bn=2an-2an+1,所以sn=b1+b2+bn=2a1-2a2+2a2-2a3+2an-2an+1=2a1-2an+1,所以sn+2an+1=2a1=1,則s2 020+2a2 021=1.2.已知正項(xiàng)數(shù)列an中,a1=1,a2=2,2an2=an-12+an+12(n2),bn=1an+an+1,數(shù)列bn的前n項(xiàng)和為sn,則s33的值是.世紀(jì)金榜導(dǎo)學(xué)號(hào)【解析】因?yàn)?an2=an-12+an+12(n2),所以數(shù)列an2是首項(xiàng)為1,公差為22-1=3的等差數(shù)列,所以an