《2024版高考数学一轮复习第六章数列课时规范练29数列求和.docx》由会员分享，可在线阅读，更多相关《2024版高考数学一轮复习第六章数列课时规范练29数列求和.docx（5页珍藏版）》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。
1、课时规范练29基础巩固组1.(2023四川成都高三期末)已知等差数列an的首项为1,公差为2,数列bn满足bn=2n-1(nN*),记数列an+bn的前n项和为Tn,则T10=()A.2 147B.1 123C.1 078D.611答案:B解析:由题意an=2n-1,an+bn=2n-1+2n-1,所以T10=10(1+19)2+1-2101-2=100+210-1=1123.故选B.2.(2023山东烟台高三月考)已知数列an满足an=nn+1,则a1+a222+a332+a202120212+a202220222=()A.20212022B.20192020C.20202021D.2022
2、2023答案:D解析:因为an=nn+1,所以ann2=1n(n+1)=1n1n+1,所以a1+a222+a332+a202120212+a202220222=1-12+1213+1314+1202112022+1202212023=1-12023=20222023.故选D.3.(2023河南商丘高三月考)数列an满足an+1=ancos n+3n,则数列an的前12项和为()A.64B.150C.108D.240答案:C解析:由已知得a2=-a1+3,a3=a2+6=-a1+9,a4=-a3+9=a1,a1+a2+a3+a4=12,由cosn周期为2,同理可得a5+a6+a7+a8=36,a
3、9+a10+a11+a12=60,故S12=12+36+60=108.故选C.4.(2022湖南岳阳高三模拟)高斯对1+2+3+100的求和运算,运用了倒序相加法的原理,此方法也称之为高斯算法.现有函数f(x)=4x4x+2,则f12023+f22023+f32023+f20222023等于()A.1 010B.1 011C.2 022D.2 023答案:B解析:因为f(x)=4x4x+2,所以f(x)+f(1-x)=4x4x+2+41-x41-x+2=4x4x+2+424x+4=1.令S=f12023+f22023+f32023+f20222023,又S=f20222023+f2021202
4、3+f20202023+f12023,两式相加得2S=12022,解得S=1011.故选B.5.(2023安徽合肥高三月考)设数列an满足a1+12a2+122a3+12n-1an=n,则数列nan的前n项和为()A.(n-1)2n-1B.(n-1)2n+1C.(n+1)2n+1-1D.(n+1)2n+1+1答案:B解析:a1+12a2+122a3+12n-1an=n,则当n2时,a1+12a2+122a3+12n-2an-1=n-1,两式相减得12n-1an=1,an=2n-1.又当n=1时,a1=1,满足an=2n-1.综上知an=2n-1,nan=n2n-1.设数列nan的前n项和为Sn
5、,Sn=120+221+322+(n-1)2n-2+n2n-1,2Sn=121+222+323+(n-1)2n-1+n2n,-Sn=1-n2n+(21+22+2n-1)=1-n2n+2(1-2n-1)1-2=(1-n)2n-1,Sn=(n-1)2n+1.故选B.6.(2023福建宁德高三月考)已知数列an满足a1=1,an-an+1=anan+1(n+1)(n+2),则an=()A.n+32n+2B.2n+2n+3C.2n+23n+1D.3n+12n+2答案:C解析:因为a1=1,an-an+1=anan+1(n+1)(n+2),所以an-an+1anan+1=1(n+1)(n+2)=1n+1
6、1n+2,即1an+11an=1n+11n+2,则1an=1an1an-1+1an-11an-2+1a21a1+1a1=1n1n+1+1n-11n+1213+11=121n+1+1=3n+12n+2(n2),所以an=2n+23n+1(n2).当n=1时,a1=2+23+1=1,符合上式.故an=2n+23n+1.故选C.7.(2023北京四中高三模拟)在数列an中,a1=1,anan+1=-2,则S100=.答案:-50解析:根据题意,由a1=1,a1a2=-2,得a2=-2;由a2a3=-2,得a3=1.所以an中所有的奇数项均为1,所有的偶数项均为-2,所以S100=a1+a2+a99+
7、a100=1-2+1-2=50(-1)=-50.8.(2023广东汕头高三月考)已知数列an的首项为-1,anan+1=-2n,则数列an的前10项和等于.答案:31解析:因为anan+1=-2n,所以an+1an+2=-2n+1,易知an0,两式相除可得an+2an=2,所以an的奇数项和偶数项均为公比为2的等比数列.又a1=-1,a2=-2a1=2,所以S10=(a1+a3+a9)+(a2+a4+a10)=-1(1-25)1-2+2(1-25)1-2=-31+231=31.9.(2022山东德州高三三模)已知数列an的前n项和为Sn,a1=3,Sn-n=12(an+1)(nN*).(1)求
8、数列an的通项公式an和前n项和Sn;(2)设bn=1(S2n+1)S2n+1(nN*),数列bn的前n项和记为Tn,证明:Tn16(nN*).(1)解:由Sn-n=12(an+1),得Sn+1-(n+1)=12(an+1+1)(nN*),两式相减可得an+1+an=2.由a1=3,得a2=-1.数列an为3,-1,3,-1,3,-1,3,即an=3,n=2k-1,-1,n=2kkN*,当n为偶数时,Sn=n23+(-1)=n;当n为奇数时,Sn=n-123+(-1)+3=n+2.故Sn=n+2,n=2k-1,n,n=2kkN*.(2)证明:因为bn=1(S2n+1)S2n+1=1(2n+1)
9、(2n+3)=1212n+112n+3,所以Tn=121315+1517+12n+112n+3=121312n+316.综合提升组10.(多选)(2022重庆高三一模)已知数列an满足an+1+ancos n=n2(nN*),则下列说法正确的是()A.a90,故a9bn3-n+12n-1,nN*.(1)解:当n=1时,a1=3;当n2时,4an-1-2Sn-1+(n-1)2-3(n-1)-4=0,nN*,所以4(an-an-1)-2an+2n-4=0,整理得an=2an-1-n+2.所以an-n=2an-1-(n-1),又a1-1=20,所以an-n0.所以an-nan-1-(n-1)=2,即
10、an-n为等比数列.所以an-n=2n,an=2n+n.(2)证明:由题意得bn+1=1+n2nbn,所以bn+1与bn同号.又因为b1=10,所以bn0,即bn+1-bn=n2nbn0,即bn+1bn.所以数列bn为递增数列,所以bnb1=1,即bn+1-bn=n2nbnn2n.当n2时,b2-b112,b3-b2222,bn-bn-1n-12n-1,累加得bn-b112+222+n-12n-1.令Tn=12+222+n-12n-1,所以12Tn=122+223+n-12n,两式相减得12Tn=12+122+123+12n-1n-12n=12(1-12n-1)1-12n-12n,所以Tn=2-n+12n-1,所以bn3-n+12n-1.又b1=1也满足bn3-n+12n-1,所以bn+1bn3-n+12n-1,nN*.