第2課時(shí)數(shù)列的遞推公式A級(jí)必備知識(shí)基礎(chǔ)練1.若數(shù)列{an}的通項(xiàng)公式為an=kn,且數(shù)列{an}是遞減數(shù)列,則實(shí)數(shù)k的取值范圍是()A.R B.(0,+∞)C.(-∞,0) D.(-∞,0]2.[2024甘肅張掖高二階段練習(xí)]在數(shù)列{an}中,a1=1,an=2an-1+1,則a5的值為()A.30 B.31C.32 D.333.[2024甘肅臨夏高二期末](多選題)下列數(shù)列{an}的通項(xiàng)公式中,是遞增數(shù)列的是()A.an=-3n-1 B.an=5n-3C.an=7+2n D.an=(-1)nn24.已知數(shù)列{an},a4=12,且滿足an+1=12an+12n,A.1 B.1C.34 D.5.已知數(shù)列{an}滿足a1>0,且an+1=nn+1an(n∈N+),則數(shù)列{an}的最大項(xiàng)是(A.a1 B.a9 C.a10 D.a116.已知數(shù)列{an}滿足a1=1,an+1=nn+2an(n∈N+),則an=(A.n+1 B.nC.2n(n7.在由火柴棒拼成的圖形中,第n個(gè)圖形由n個(gè)正方形組成.通過觀察可以發(fā)現(xiàn):第n個(gè)圖形中火柴棒的個(gè)數(shù)an與第n+1個(gè)圖形中火柴棒的個(gè)數(shù)an+1之間的關(guān)系是.

8.已知數(shù)列{an}的通項(xiàng)公式為an=kn2n+3(k∈(1)當(dāng)k=1時(shí),判斷數(shù)列{an}的單調(diào)性;(2)若數(shù)列{an}是遞減數(shù)列,求實(shí)數(shù)k的取值范圍.B級(jí)關(guān)鍵能力提升練9.若數(shù)列{an}滿足a2n=a2n-1+a2n+1(n∈N+),則稱{an}為“Y型數(shù)列”,則下列數(shù)列不可能是“Y型數(shù)列”的是()A.-1,0,1,0,-1,0,1,…B.1,2,1,3,5,2,3,…C.0,0,0,0,0,0,0,…D.2,1,-1,0,1,2,1,…10.(多選題)若數(shù)列{an}是遞增數(shù)列,則數(shù)列{an}的通項(xiàng)公式可以為()A.an=1n B.an=n2C.an=1-2n D.an=2n+111.[2024甘肅白銀高二期中]在數(shù)列{an}中,若a1=2,an=1-1an-1(n≥2),則a2024A.-1 B.1C.2 D.112.已知數(shù)列an=n-34n-35,則數(shù)列{an}的前A.a1,a30 B.a30,a1 C.a5,a6 D.a6,a513.已知an=n2-tn+2024(n∈N+,t∈R),若數(shù)列{an}中的最小項(xiàng)為第3項(xiàng),則t的取值范圍為.

14.請寫出一個(gè)符合要求①②③的數(shù)列{an}的通項(xiàng)公式.①{an}為無窮數(shù)列;②{an}為遞增數(shù)列;③0<an<2.這個(gè)數(shù)列的通項(xiàng)公式可以是.

15.若數(shù)列(2n-1)56n中的最大項(xiàng)是第k項(xiàng),則k=.

16.在數(shù)列{an}中,a1=1,an+1=anan+1(n∈(1)試寫出數(shù)列1an的一個(gè)遞推關(guān)系;(2)求數(shù)列{an}的通項(xiàng)公式.C級(jí)學(xué)科素養(yǎng)創(chuàng)新練17.[2024甘肅白銀高二期中]若數(shù)列{an}不是遞增數(shù)列,但數(shù)列{|an|}是遞增數(shù)列,則稱{an}是T數(shù)列.下列數(shù)列不是T數(shù)列的是()A.{22-2n} B.{(-4)n}C.1-n22n D.18.在數(shù)列{an}中,a1=2,且an+1=an+ln1+1n,求數(shù)列{an}的通項(xiàng)公式.

第2課時(shí)數(shù)列的遞推公式1.C∵{an}是遞減數(shù)列,∴an+1-an=k(n+1)-kn=k<0.2.B∵在數(shù)列{an}中,a1=1,an=2an-1+1,∴a2=2×1+1=3,a3=2×3+1=7,a4=2×7+1=15,a5=2×15+1=31.故選B.3.BC對于A,∵an+1-an=-3(n+1)-1+3n+1=-3<0,∴數(shù)列{an}為遞減數(shù)列,故A錯(cuò)誤;對于B,∵an+1-an=5(n+1)-3-5n+3=5>0,∴數(shù)列{an}為遞增數(shù)列,故B正確;對于C,∵an+1-an=7+2n+1-7-2n=2n>0,∴數(shù)列{an}為遞增數(shù)列,故C正確;對于D,an+1-an=(-1)n+1(n+1)2-(-1)nn2=(-1)n+1(2n2+2n+1),∵2n2+2n+1=2(n+12)2+12>0,∴當(dāng)n∴數(shù)列{an}不是遞增數(shù)列,故D錯(cuò)誤.故選BC.4.A由a4=12,且an+1=12an+12n,得a4=12a3+123,可得a3=34.由a3=12a2+122,可得a2=1.由a2=15.A因?yàn)閍1>0且an+1=nn+1a所以an>0,an+1an=nn+1<1,所以an+1<a6.C由題意,an+1=nn+2an,即an+1·(n+2)(n+1)=an·(n+1)因?yàn)閍1=1,所以an+1·(n+2)(n+1)=an·(n+1)n=…=a1·2·1=2,可得an=2n(n+17.an+1-an=3(n≥1)通過觀察,第1個(gè)圖形中,火柴棒有4根;第2個(gè)圖形中,火柴棒有4+3根;第3個(gè)圖形中,火柴棒有4+3+3=4+3×2根;第4個(gè)圖形中,火柴棒有4+3+3+3=4+3×3根.可以發(fā)現(xiàn),從第二項(xiàng)起,每一項(xiàng)與前一項(xiàng)的差都等于3,即a2-a1=3,a3-a2=3,a4-a3=3,…,an+1-an=3(n≥1).8.解(1)當(dāng)k=1時(shí),an=n2n+3,所以an+1所以an+1-an=n+12故數(shù)列{an}是遞增數(shù)列.(2)若數(shù)列{an}是遞減數(shù)列,則an+1-an<0恒成立,即an+1-an=kn+k2n+5-kn2n+3=3k(2n+5)(2n+3)<0恒成立.因?yàn)?29.B數(shù)列{an}滿足a2n=a2n-1+a2n+1(n∈N+),即數(shù)列的每個(gè)偶數(shù)項(xiàng)都等于其相鄰兩項(xiàng)的和,B不符合條件,故選B.10.BDan=1n,a1=1,a2=12,則{an}不是遞增數(shù)列,因此Aan=n2+n,an-an-1=n2+n-(n-1)2-(n-1)=2n>0,則{an}是遞增數(shù)列,因此B符合題意;an=1-2n,an-an-1=(1-2n)-[1-2(n-1)]=-2,則{an}不是遞增數(shù)列,因此C不符合題意;an=2n+1,函數(shù)y=2x+1為增函數(shù),則{an}是遞增數(shù)列,因此D符合題意.故選BD.11.B由題意得a1=2,a2=1-1a1=1-12=12,a3=1-1a2=1-2=-1,a4=1-故{an}中的項(xiàng)以3為周期重復(fù)出現(xiàn),故a2024=a674×3+2=a2=12.故選B12.Dan=n-34n-35=n-35+35-34n-35=1+35-34n-35,又因?yàn)閒(n)=35當(dāng)n=6時(shí),an最大,即最大項(xiàng)為a6.故選D.13.(5,7)已知an=n2-tn+2024(n∈N+,t∈R),∵數(shù)列{an}中最小項(xiàng)為第3項(xiàng),∴52<t2<714.an=2-1n(答案不唯一15.6根據(jù)題意知,當(dāng)k>1時(shí),有a即(解得112<k<13又k∈N+,所以k=6.又a1=56<a6,所以k=616.解(1)因?yàn)閍1=1,an+1=an所以an≠0,1an所以1an+1=1an+1(n≥1),1a(2)由(1)可得1an+1-1an=1(n≥1),則1a2-1a1=1,1將上述(n-1)個(gè)等式相加,得1an-1a1=n-1,即1an=n,即an=1n,n≥2.當(dāng)n=1時(shí),a1=11=1,符合上式,所以a17.D當(dāng)an=22-2n時(shí),{an}是遞減數(shù)列,|an|=22-2,n=1,2n-22,n≥2,因?yàn)閨a1|-|a2|=42-6<0,當(dāng)n≥2時(shí),{2n-22}是遞增數(shù)列,所以{|an|}是遞增數(shù)列當(dāng)an=(-4)n時(shí),易知{(-4)n}不是遞增數(shù)列,因?yàn)閨(-4)n|=4n,所以{|an|}是遞增數(shù)列,所以{(-4)n}是T數(shù)列,故B錯(cuò)誤;因?yàn)?-n22n=121n-n,所以1-n22n是遞減數(shù)列,因?yàn)?21n-n=12n-1n≥0,且{|