2024-2025學(xué)年新教材高中數(shù)學(xué)第五章數(shù)列5.1.2數(shù)列中的遞推課時(shí)素養(yǎng)評(píng)價(jià)含解析新人教B版選擇性必修第三冊(cè)

PAGE“課時(shí)素養(yǎng)評(píng)價(jià)”二數(shù)列中的遞推(25分鐘·50分)一、選擇題(每小題5分,共20分,多選題全部選對(duì)的得5分,選對(duì)但不全的得3分,有選錯(cuò)的得0分)1.數(shù)列QUOTE,QUOTE,QUOTE,QUOTE,…的遞推公式可以是 ()A.an=QUOTE(n∈N+)B.an=QUOTE(n∈N+)C.a1=QUOTE,an+1=QUOTEan(n∈N+)D.a1=QUOTE,an+1=2an(n∈N+)【解析】選C.數(shù)列從第2項(xiàng)起,后一項(xiàng)是前一項(xiàng)的QUOTE,故遞推公式為a1=QUOTE,an+1=QUOTEan(n∈N+).2.數(shù)列{an}中,an+1=an+2-an,a1=2,a2=5,則a5= ()A.-3 B.-11 C.-5 D.19【解析】選D.a3=a2+a1=5+2=7,a4=a3+a2=7+5=12,a5=a4+a3=12+7=19.3.(多選題)假如{an}為遞增數(shù)列,則{an}的通項(xiàng)公式可以為 ()A.an=2n+3 B.an=-n2-3n+1C.an=QUOTE D.an=1+log2n【解析】選AD.A是n的一次函數(shù),一次項(xiàng)系數(shù)為2,所以為遞增數(shù)列;B是n的二次函數(shù),二次項(xiàng)系數(shù)為-1,且對(duì)稱軸為n=-QUOTE,所以為遞減數(shù)列;C是n的指數(shù)函數(shù),且底數(shù)為QUOTE,是遞減數(shù)列;D是n的對(duì)數(shù)函數(shù),且底數(shù)為2,是遞增數(shù)列.【加練·固】在遞減數(shù)列{an}中,an=kn(k為常數(shù)),則實(shí)數(shù)k的取值范圍是 ()A.R B.(0,+∞)C.(-∞,0) D.(-∞,0]【解析】選C.因?yàn)閧an}是遞減數(shù)列,所以an+1-an=k(n+1)-kn=k<0.4.已知數(shù)列{an},a1=1,lnan+1-lnan=1,則數(shù)列{an}的通項(xiàng)公式是 ()A.an=n B.an=QUOTEC.an=en-1 D.an=QUOTE【解析】選C.因?yàn)閘nan+1-lnan=1,所以lnQUOTE=1.所以QUOTE=e.由累乘法可得an=en-1.二、填空題(每小題5分,共10分)5.已知數(shù)列{an}滿意a1=1,an=nan-1(n≥2),則a5=________.

【解析】因?yàn)閍n=nan-1,且n≥2,所以當(dāng)n=2時(shí),a2=2a1=2;當(dāng)n=3時(shí),a3=3a2=6;當(dāng)n=4時(shí),a4=4a3=24;當(dāng)n=5時(shí),a5=5a4=120.故a5=120.答案:1206.數(shù)列{an}中,a1=1,對(duì)全部的n≥2,都有a1·a2·a3·…·an=n2,則a3+a5=________.

【解析】由題意a1a2a3=32,a1a2=22,a1a2a3a4a5=52,a1a2a3a4=42,則a3=QUOTE=QUOTE,a5=QUOTE=QUOTE.故a3+a5=QUOTE.答案:QUOTE三、解答題(每小題10分,共20分)7.依據(jù)下列條件,寫出數(shù)列的前四項(xiàng),并歸納猜想它的通項(xiàng)公式.(1)a1=1,an+1=an+QUOTE(n∈N+).(2)a1=2,a2=3,an+2=3an+1-2an(n∈N+).【解析】(1)a1=1,a2=QUOTE,a3=QUOTE=2,a4=QUOTE.猜想an=QUOTE.(2)a1=2,a2=3,a3=5,a4=9.猜想an=2n-1+1.8.已知數(shù)列{an}的前n項(xiàng)和Sn=2n2-3n+1,求{an}的通項(xiàng)公式an.【解析】因?yàn)閿?shù)列{an}的前n項(xiàng)和Sn=2n2-3n+1,所以a1=S1=2×12-3×1+1=0,當(dāng)n≥2時(shí),an=Sn-Sn-1=(2n2-3n+1)-[2(n-1)2-3(n-1)+1]=4n-5.n=1時(shí),4n-5=-1≠a1,所以{an}的通項(xiàng)公式an=QUOTE(n∈N+).(15分鐘·30分)1.(5分)已知,在數(shù)列{an}中,a1=3,a2=6,且an+2=an+1-an,則a2012= ()A.3 B.-3 C.6 D.-6【解析】選C.由題意知:a3=a2-a1=3,a4=a3-a2=-3,a5=a4-a3=-6,a6=a5-a4=-3,a7=a6-a5=3,a8=a7-a6=6,a9=a8-a7=3,a10=a9-a8=-3,…故知{an}是周期為6的數(shù)列,所以a2012=a2=6.2.(5分)已知數(shù)列{an}滿意an=QUOTE若對(duì)于隨意的n∈N+都有an<an+1成立,則實(shí)數(shù)a的取值范圍是 ()A.(1,4) B.(2,5) C.(1,6) D.(4,6)【解析】選A.因?yàn)閷?duì)于隨意的n∈N+都有an<an+1成立,所以數(shù)列{an}單調(diào)遞增,所以應(yīng)滿意QUOTE解得1<a<4.故實(shí)數(shù)a的取值范圍是(1,4).3.(5分)已知Sn是數(shù)列{an}的前n項(xiàng)和,且滿意Sn=n2+n(n∈N+),則S3=________,數(shù)列{an}的通項(xiàng)公式an=________.

【解析】由Sn=n2+n,所以S3=9+3=12.當(dāng)n=1時(shí),a1=S1=1+1=2,當(dāng)n≥2時(shí),an=Sn-Sn-1=n2+n-(n-1)2-(n-1)=2n,當(dāng)n=1時(shí),得a1=2成立,所以an=2n.答案:122n4.(5分)若數(shù)列{an}的通項(xiàng)公式為an=-2n2+13n,關(guān)于該數(shù)列,有以下四種說法:(1)該數(shù)列有無限多個(gè)正數(shù)項(xiàng).(2)該數(shù)列有無限多個(gè)負(fù)數(shù)項(xiàng).(3)該數(shù)列的最大項(xiàng)就是函數(shù)f(x)=-2x2+13x的最大值.(4)-70是該數(shù)列中的一項(xiàng).其中正確說法的序號(hào)為________.

【解析】令-2n2+13n>0,得0<n<QUOTE,故數(shù)列{an}有6項(xiàng)是正數(shù)項(xiàng),有無限個(gè)負(fù)數(shù)項(xiàng),所以(1)錯(cuò)誤,(2)正確;當(dāng)n=3時(shí),數(shù)列{an}取到最大值,而當(dāng)x=3.25時(shí),函數(shù)f(x)取到最大值,所以(3)錯(cuò)誤;令-2n2+13n=-70,得n=10,或n=-QUOTE(舍去).即-70是該數(shù)列的第10項(xiàng),所以(4)正確.答案:(2)(4)5.(10分)已知數(shù)列{an}滿意a1=QUOTE,n≥2時(shí),anan-1=an-1-an,求數(shù)列{an}的通項(xiàng)公式.【解析】因?yàn)閍nan-1=an-1-an,所以QUOTE-QUOTE=1.所以n≥2時(shí),QUOTE=QUOTE+QUOTE+QUOTE+…+QUOTE=2+QUOTE=n+1.所以QUOTE=n+1,所以當(dāng)n≥2時(shí),an=QUOTE.當(dāng)n=1時(shí),a1=QUOTE也適合上式,所以an=QUOTE(n∈N+).1.(2024·辛集高二檢測(cè))已知an=QUOTE(n∈N+),則數(shù)列{an}的前50項(xiàng)中最小項(xiàng)和最大項(xiàng)分別是 ()A.a1,a50 B.a1,a44C.a45,a50 D.a44,a45【解析】選D.an=QUOTE=1+QUOTE,因?yàn)?42=1936,452=2025,所以n≤44時(shí),數(shù)列{an}單調(diào)遞減,且0<an<1;n≥45時(shí),數(shù)列{an}單調(diào)遞減,且an>1.所以在數(shù)列{an}的前50項(xiàng)中最小項(xiàng)和最大項(xiàng)分別是a44,a45.2.設(shè)an=n2-2kn+6(n∈N+,k∈R)(1)證明:k≤1是{an}為遞增數(shù)列的充分不必要條件;(2)若對(duì)隨意的n∈N+,QUOTE≥1,求k的取值范圍.【解析】(1)an+1-an=(n+1)2-2k(n+1)+6-(n2-2k