/課時規(guī)范練26《素養(yǎng)分級練》P365基礎鞏固組1.(2023·河南平頂山高三月考)已知數(shù)列{an}的前n項和為Sn,且Sn=2n+1,則a10=()A.512 B.1025C.256 D.1024答案:A解析:由數(shù)列{an}的前n項和為Sn,且Sn=2n+1,得a10=S10-S9=(210+1)-(29+1)=512.故選A.2.(2023·廣東佛山高三月考)已知數(shù)列{an}滿足a1=1,an+1=an4an+1(n∈N*),則滿足an>1A.7 B.8 C.9 D.10答案:C解析:因為a1=1,an+1=an4an+1,所以1an+1=4+1an,即1an+1?1an=4.又1a1=1,所以數(shù)列1an是以1為首項,4為公差的等差數(shù)列,于是1an=1+4(n-1)=4n-3,所以an=14n-3.由an>137,即13.(2023·山東東營高三月考)在數(shù)列{an}中,a1=2,an=1-1an-1(n≥2),則aA.-12 B.12 C.-1答案:C解析:由a1=2,an=1-1an-1(n≥2),可得a2=1-1a1=12,a3=1-1a2=-1,a4=1-1a3=2,a5=1-1a4=12,故數(shù)列{an}為周期為3的周期數(shù)列,而4.(2023·浙江溫州高三模擬)已知數(shù)列{an}為遞增數(shù)列,前n項和Sn=n2+n+λ,則實數(shù)λ的取值范圍是()A.(-∞,2] B.(-∞,2)C.(-∞,0] D.(-∞,0)答案:B解析:當n≥2時,an=Sn-Sn-1=n2+n+λ-[(n-1)2+(n-1)+λ]=2n,可知當n≥2時,{an}是遞增數(shù)列,因此要使{an}為遞增數(shù)列只需滿足a2>a1,即4>2+λ?λ<2.故選B.5.(2023·山東濟南高三模擬)在數(shù)列{an}中,a1=5,a2=9,若數(shù)列{an+n2}是等差數(shù)列,則{an}的最大項的值為()A.9 B.11 C.454 答案:B解析:令bn=an+n2,∵a1=5,a2=9,∴b2=a2+4=13,b1=a1+1=6,∴數(shù)列{an+n2}的公差為13-6=7,則an+n2=6+7(n-1)=7n-1,∴an=-n2+7n-1=-n-722+454.又n∈N*,∴當n=3或4時,an取最大值-14+454=116.(多選)(2023·福建寧德高三模擬)已知數(shù)列{an}滿足a1=1,an-an+1=nanan+1,則下列說法正確的有()A.數(shù)列1an為等差數(shù)列B.a3=1C.an=2D.數(shù)列{an}的最大項的值為1答案:BCD解析:an-an+1=nanan+1,等式兩邊同除以anan+1,得1an+1?1an=n,因此數(shù)列1an不是等差數(shù)列,故A錯誤;又1a2?1a1=1,1a3?1a2=2,…,1an?1an-1=n-1,n≥2,累加可得1a2?1a1+1a3?1a2+…+1an?1an-1=1+2+…+n-1,即1an?1a1=(n-1)n2,又a1=1,所以1an=(n-1)n7.已知數(shù)列{an}的前n項和為Sn,且滿足Sn+an=4,則S4=.

答案:15解析:當n=1時,有2a1=4,可得a1=2.當n≥2時,由Sn+an=4可得Sn-1+an-1=4,兩式作差得2an-an-1=0,所以anan-1=12,即數(shù)列{an}是以2為首項,18.(2023·湖南師大附中高三期中)已知在數(shù)列{an}中,a1=2,a1+a22+a33+…+ann=an+答案:2n解析:a1+a22+a33+…+ann=an+1-2,當n≥2時,a1+a22+a33+…+an-1n-1=an-2,則ann=an+1-an,即an+1n+1=ann.當n=1時,a1=a9.(2023·山東濰坊高三模擬)數(shù)列{an}滿足an+1=5an+3×5n+1,且a1=6,則數(shù)列{an}的通項公式為.

答案:an=3n-95·5n解析:因為an+1=5an+3×5n+1,所以an+15n+1=an5n+3,即an+15n+1?an5n=3,所以an5n是等差數(shù)列,而a15=綜合提升組10.已知等差數(shù)列{an},其前n項和為Sn,若a1=13,S5=45,則nSn的最大值為()A.400 B.405 C.410 D.415答案:B解析:設等差數(shù)列{an}的公差為d,則S5=5a1+5×42d=5a1+10d=45,解得d=-2,所以Sn=na1+n(n-1)d2=13n-n(n-1)=14n-n2,則nSn=14n2-n3.令bn=14n2-n3,則bn+1-bn=[14(n+1)2-(n+1)3]-(14n2-n3)=-3n2+25n+13,所以當n≤8時,bn+1-bn>0,即b1<b2<…<b9;當n≥9時,bn+1-bn<0,即b9>b10>…,所以數(shù)列{bn}中的最大項為b9=14×92-11.(2023·安徽蚌埠高三期中)已知數(shù)列{an}的首項a1=2,且滿足an+1=an+12n(n∈N*).若對于任意的正整數(shù)n,存在M使得an<M恒成立,則M的最小值是.

答案:3解析:由已知得an+1-an=12n,∴當n=1時,a2-a1=121,當n=2時,a3-a2=122,當n=3時,a4-a3=123,……,當n=n-1時,an-an-1=12n-1(n≥2),以上各式相加得an-a1=121+122+123+…+12n-1=12×[1-(12)

n-1]1-12=1-12n-1,n≥2.又a1=2,∴an=3-12n-1,n≥2,又a1=2也符合上式,故an=3-12n-1.∵12n-1>0,∴an<3.若對于任意的正整數(shù)n,存在M使得an<M恒成立,則有M≥3,故M創(chuàng)新應用組12.(2023·湖南長沙高三期中)數(shù)列{an}的前n項的和Sn滿足Sn+1+Sn=n(n∈N*),則下列選項正確的是()A.數(shù)列{an+1+an}是常數(shù)列B.若a1<13,則{anC.若a1=-1,則S2022=1013D.若a1=1,則{an}的最小項的值為-1答案:D解析:當n=1時,S2+S1=2a1+a2=1,當n≥2時,Sn+Sn-1=n-1,則an+1+an=1,而a1+a2=1不一定成立,故{an+1+an}不一定是常數(shù)列,故A錯誤;由an+1+an=an+an-1=…=a3+a2=1,得an+1=an-1=an-3=…且an=an-2=an-4=…,即{an}不是單調數(shù)列,故B錯誤;若a1=-1,則a2=3,a3=-2,故當n≥2時,{an}的偶數(shù)項的值為3,奇數(shù)項的值為-2,而S2022=a1+(a2+a3)+(a4+a5)+…+(a2020+a2021)+a2022=-1+1010+3=1012,故C錯誤;若a1=1,則a2=-1,a3=2,故當n≥2時,{an}的偶數(shù)項的值為-1,奇數(shù)項的值為2,故{an}的最小項的值為-1,故D正確.故選D.13.(2023·遼寧錦州高三月考)已知數(shù)列{a