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
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.20222023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,a9+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+42×4x+4=1.令S=f12023+f22023+f32023+f20222023,又S=f20222023+f20212023+f20202023+f12023,两式相加得2S=1×2022,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=n·2n-1.设数列nan的前n项和为Sn,Sn=1×20+2×21+3×22+(n-1)·2n-2+n·2n-1,2Sn=1×21+2×22+3×23+(n-1)·2n-1+n·2n,-Sn=1-n·2n+(21+22+2n-1)=1-n·2n+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+11n+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,an·an+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+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+2×31=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:Tn<16(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)(2n+3)=1212n+112n+3,所以Tn=121315+1517+12n+112n+3=121312n+3<16.综合提升组10.(多选)(2022·重庆高三一模)已知数列an满足an+1+an·cos n=n2(nN*),则下列说法正确的是()A.a9<a10B.a9+a11=20C.数列an的前10项和为定值D.数列an的前20项和为定值答案:AD解析:取n=9得a10-a9=92>0,故a9<a10,故A正确;取n=10得a11+a10=102,又a10-a9=92,两式相减得a11+a9=19,故B不正确;由题知,a2n-a2n-1=(2n-1)2,a2n+1+a2n=(2n)2,a2n+2-a2n+1=(2n+1)2,-得a2n+1+a2n-1=4n-1,+得a2n+2+a2n=8n2+4n+1,所以(a3+a5)+(a7+a9)+(a4+a6)+(a8+a10)为定值,题中条件只限制a2-a1=1,所以a1+a2的值不确定,故前10项和无法确定,故C不正确;前20项中奇数项有10项,相邻两项的和确定,故这10项的和确定,同理10个偶数项的和确定,故前20项和为定值,故D正确.故选AD.11.(2023·湖北天门高三期中)已知数列an的前n项和为Sn,且满足an+an+1+an+2=cosn3,a1=1,则S2 023=. 答案:1解析:S2023=a1+(a2+a3+a4)+(a5+a6+a7)+(a2021+a2022+a2023)=1+cos23+cos53+cos20183+cos20213=1+337×cos23+cos53=1.12.(2023·辽宁沈阳高三月考)已知数列pn是首项为2的等差数列,qn是公比为2的等比数列,且满足p3=q2,p7=q3.设数列an满足an=pn·qn.(1)求an的通项公式;(2)在bn=nn+1an,bn=log4ann+2,bn=(n+3)·2nan+1an这三个条件中任选一个补充在下面横线上,并加以解答.已知数列bn满足,求bn的前n项和Tn. 解:(1)设数列pn的公差为d,数列qn的首项为q1.由题意得2+2d=2q1,2+6d=4q1,解得d=1,q1=2,则pn=n+1,qn=2n,所以an=(n+1)·2n.(2)若选bn=nn+1an,即bn=nn+1·(n+1)·2n=n·2n,所以Tn=1×21+2×22+3×23+n×2n,则2Tn=1×22+2×23+3×24+n×2n+1,两式相减得-Tn=21+22+23+2n-n×2n+1=2-2n+11-2-n×2n+1=(1-n)2n+1-2,所以Tn=(n-1)×2n+1+2.若选bn=log4ann+2,即bn=log4n+1n+2+log42n=log4n+1n+2+n2,所以Tn=log423+log434+log4n+1n+2+12(1+2+n)=log423×34××n+1n+2+(1+n)n4=log42n+2+(1+n)n4.若选bn=(n+3)·2nan+1an,则bn=(n+3)·2nan+1an=1an1an+1,所以Tn=1a11a2+1a21a3+1an1an+1=1a11an+1=141(n+2)2n+1.创新应用组13.(2023·江苏扬州高三月考)已知数列an,当n2k-1,2k)时,an=k(kN*),则数列an的前2n项的和为. 答案:(n-1)·2n+n+2解析:当am=k=1时,m1,2),共2-1=1项;当am=k=2时,m2,4),共4-2=2项;当am=k=3时,m4,8),共8-4=4项;当am=k=n时,m2n-1,2n),共2n-2n-1=2n-1项.又因为a2n=n+1,所以,数列an的前2n项的和为1·20+2·21+3·22+n·2n-1+n+1,记S=1·20+2·21+3·22+n·2n-1,则2S=1·21+2·22+(n-1)·2n-1+n·2n,两式作差可得-S=1+2+22+2n-1-n·2n=1-2n1-2-n·2n=(1-n)·2n-1,所以S=(n-1)·2n+1,因此,数列an的前2n项的和为S+n+1=(n-1)·2n+n+2.14.(2023·浙江镇海高三模拟)已知数列an的前n项和Sn满足4an-2Sn+n2-3n-4=0,nN*.数列bn满足b1=1,2nbn+1=anbn,nN*.(1)求证:数列an-n为等比数列,并求数列an的通项公式;(2)求证:bn+1>bn3-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=1>0,所以bn>0,即bn+1-bn=n2nbn>0,即bn+1>bn.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+1>bn3-n+12n-1,nN*.
