PAGEPAGE1第8講數(shù)列求和1.已知數(shù)列{an}的前n項和Sn=n2+3n+2,則{an}的通項公式為()A.an=2n-2 B.an=2n+2C.an=6,n=122.已知數(shù)列{an}中,a1=a2=1,an+2=an+2,A.1121 B.1122C.1123 D.11243.已知數(shù)列{an}滿意an+1-an=2,a1=-5,則|a1|+|a2|+…+|a6|=()A.9 B.15 C.18 D.304.設(shè){an}是公差為2的等差數(shù)列,bn=a2n.若{bn}為等比數(shù)列,則b1+b2+b3+b4+bA.142 B.124 C.128 D.1445.已知等差數(shù)列{an}的公差為d,關(guān)于x的不等式dx2+2a1x≥0的解集為[0,9],則使數(shù)列{an}的前n項和Sn最大的正整數(shù)n的值是()A.4 B.5 C.6 D.76.(2024山西八校第一次聯(lián)考)已知數(shù)列{an}滿意:a1=1,an+1=anan+2(n∈N*),若bn+1=(n-λ)1aA.(2,+∞) B.(3,+∞) C.(-∞,2) D.(-∞,3)7.設(shè)數(shù)列{an}的前n項和為Sn,且Sn=a1(4n-1)8.設(shè)等差數(shù)列{an}的前n項和為Sn,且a3=3,S4=10,則∑k=1n9.我國古代數(shù)學(xué)名著《九章算術(shù)》有如下問題:“今有女善織,日自倍,五日織五尺,問:日織幾何?”意思是“一女子擅長織布,每天織的布都是前一天的2倍,已知她5天織了5尺布,問這女子每天分別織多少布?”依據(jù)題目的條件,若要使織布的總尺數(shù)不少于30尺,該女子所需的天數(shù)至少為天.

10.已知數(shù)列{xn}的各項均為正整數(shù),且滿意xn+1=xn2,xn為偶數(shù),xn+1,xn為奇數(shù),11.(2024鄭州第一次質(zhì)檢)已知等差數(shù)列{an}的前n項和為Sn,且a2+a5=25,S5=55.(1)求數(shù)列{an}的通項公式;(2)設(shè)anbn=13n-1,求數(shù)列{b12.(2024武漢武昌調(diào)研考試)已知數(shù)列{an}的前n項和Sn=2an-2.(1)求數(shù)列{an}的通項公式;(2)令bn=an·log2an,求數(shù)列{bn}的前n項和Tn.13.已知數(shù)列{an}滿意a1=1,a1+12a2+13a3+…+1nan=an+1-1(n∈N*),數(shù)列{an(1)求數(shù)列{an}的通項公式;(2)設(shè)bn=1Sn,Tn是數(shù)列{bn}的前n項和,求使得Tn<m10對全部

答案全解全析1.Da1=S1=6,當(dāng)n≥2時,an=Sn-Sn-1=2n+2,∴an=62.C由題意可知,數(shù)列{a2n}是首項為1,公比為2的等比數(shù)列,數(shù)列{a2n-1}是首項為1,公差為2的等差數(shù)列,故數(shù)列{an}的前20項和為1×(1-3.C由an+1-an=2可得,數(shù)列{an}是等差數(shù)列,公差d=2.又a1=-5,所以an=2n-7.所以|a1|+|a2|+|a3|+|a4|+|a5|+|a6|=5+3+1+1+3+5=18.4.B由題意,得an=a1+(n-1)×2=a1+2n-2.∵{bn}為等比數(shù)列,∴b22=b1·b3.∴(a4)2=a2·a∴(a1+6)2=(a1+2)(a1+14),解得a1=2.∴an=2+2(n-1)=2n.∴bn=a2n=2n+1.∴b1+b2+b3+b4+b5=22+23+24+25+25.B∵關(guān)于x的不等式dx2+2a1x≥0的解集為[0,9],∴0,9是一元二次方程dx2+2a1x=0的兩個實數(shù)根,且d<0.∴-2a1d=9,a1=-9d2.∴an=a1+(n-1)d=n-112d,可得a5=-126.C由an+1=anan+2,知1an+1=2an+1,即1an+1+1=21an+1,所以數(shù)列1an+1是首項為1a1+1=2,公比為2的等比數(shù)列.所以1a7.答案12解析a4=32,即S4-S3=32.因為Sn=a1(所以255a13所以a1=128.答案2n答案設(shè)公差為d,則a1+2d=3,4a1∴前n項和Sn=1+2+…+n=n(∴1Sn=2n∴∑k=1n1Sk=21-12+12-13+…+19.答案8解析設(shè)該女子第1天織布x尺,則x(1-2所以前n天所織布的尺數(shù)為531(2n令531(2n-1)≥30,則2n≥187.又n∈N*故n的最小值為8.10.答案{1,2,3,4,8}解析由題意,得x3=1,x4=2或x3=2,x4=1.當(dāng)x3=1時,x2=2,從而x1=1或4;當(dāng)x3=2時,x2=1或4.因此,當(dāng)x2=1時,x1=2;當(dāng)x2=4時,x1=8或3.綜上,x1全部可能取值的集合為{1,2,3,4,8}.11.解析(1)設(shè)等差數(shù)列{an}的公差為d.由題意得2a1+5d=25,5a1+(2)由anbn=13n-1,得bn=1a∴Tn=b1+b2+…+bn=1=1=n212.解析(1)當(dāng)n=1時,a1=2a1-2,所以a1=2.當(dāng)n≥2時,Sn-1=2an-1-2,Sn-Sn-1=(2an-2)-(2an-1-2),即an=2an-1.所以數(shù)列{an}是以2為首項,2為公比的等比數(shù)列.所以an=2n.(2)由(1),得bn=2nlog22n=n·2n,所以Tn=1×21+2×22+3×23+…+(n-1)×2n-1+n×2n,2Tn=1×22+2×23+3×24+…+(n-1)×2n+n×2n+1.兩式相減,得-Tn=21+22+23+…+2n-n×2n+1=2(1=(1-n)2n+1-2.所以Tn=(n-1)2n+1+2.13.解析(1)∵a1+12a2+…+1nan=an+1-1(n∈N*∴當(dāng)n≥2時,a1+12a2+…+1n-1a兩式相減,得1nan=an+1-an(n≥2,n∈N*即an+1an=又a2a1=1+a1∴an+1an=∴當(dāng)n≥2時,an=anan-