考點(diǎn)規(guī)范練35數(shù)學(xué)歸納法基礎(chǔ)鞏固組1.用數(shù)學(xué)歸納法證明2n>2n+1,n的第一個(gè)取值應(yīng)是()A.1B.2C.3D.4答案C解析當(dāng)n=1時(shí),21=2,2×1+1=3,2n>2n+1不成立;當(dāng)n=2時(shí),22=4,2×2+1=5,2n>2n+1不成立;當(dāng)n=3時(shí),23=8,2×3+1=7,2n>2n+1成立.故n的第一個(gè)取值應(yīng)是3.2.已知f(n)=1n+1n+1+1n+2+1n2,則()A.f(n)中共有n項(xiàng),當(dāng)n=2時(shí),f(2)=12+13B.f(n)中共有n+1項(xiàng),當(dāng)n=2時(shí),f(2)=12+13C.f(n)中共有n2-n項(xiàng),當(dāng)n=2時(shí),f(2)=12+13D.f(n)中共有n2-n+1項(xiàng),當(dāng)n=2時(shí),f(2)=12+13+14答案D解析總項(xiàng)數(shù)為n2-(n-1),f(2)=12+13+14.故選D.3.用數(shù)學(xué)歸納法證明“1+a+a2+an+1=1-an+21-a(a1,nN*)”,在驗(yàn)證n=1時(shí),左端計(jì)算所得的結(jié)果是()A.1B.1+aC.1+a+a2D.1+a+a2+a3答案C解析當(dāng)n=1時(shí),左邊=1+a+a2.故選C.4.某個(gè)命題與自然數(shù)n有關(guān),若n=k(kN*)時(shí)命題成立,則可推得當(dāng)n=k+1時(shí)該命題也成立.現(xiàn)已知當(dāng)n=5時(shí),該命題不成立,則可推得()A.當(dāng)n=6時(shí),該命題不成立B.當(dāng)n=6時(shí),該命題成立C.當(dāng)n=4時(shí),該命題不成立D.當(dāng)n=4時(shí),該命題成立答案C解析因?yàn)楫?dāng)n=k時(shí)命題成立可推出當(dāng)n=k+1時(shí)成立,所以當(dāng)n=5時(shí)命題不成立,則當(dāng)n=4時(shí)命題也一定不成立.5.對(duì)于不等式n2+n<n+1(nN*),某同學(xué)用數(shù)學(xué)歸納法證明的過(guò)程如下:(1)當(dāng)n=1時(shí),12+1<1+1,不等式成立.(2)假設(shè)當(dāng)n=k(kN*)時(shí),不等式成立,即k2+k<k+1,則當(dāng)n=k+1時(shí),(k+1)2+(k+1)=k2+3k+2<(k2+3k+2)+(k+2)=(k+2)2=(k+1)+1.所以當(dāng)n=k+1時(shí),不等式成立,則上述證法()A.過(guò)程全部正確B.n=1驗(yàn)得不正確C.歸納假設(shè)不正確D.從n=k到n=k+1的推理不正確答案D解析在n=k+1時(shí),沒(méi)有應(yīng)用n=k時(shí)的假設(shè),不是數(shù)學(xué)歸納法.6.在數(shù)列an中,已知a1=2,an+1=an3an+1(nN*),依次計(jì)算出a2,a3,a4的值分別為;歸納可知an=. 答案27,213,21926n-5解析a1=2,a2=23×2+1=27,a3=273×27+1=213,a4=2133×213+1=219.由此,猜想an的分子為2,分母是以1為首項(xiàng),6為公差的等差數(shù)列.故an=26n-5.用數(shù)學(xué)歸納法可證明.7.用數(shù)學(xué)歸納法證明:12+22+n2+22+12=n(2n2+1)3,第二步證明由“k到k+1”時(shí),左邊應(yīng)加. 答案(k+1)2+k2解析當(dāng)n=k時(shí),左邊=12+22+k2+22+12,當(dāng)n=k+1時(shí),左邊=12+22+k2+(k+1)2+k2+22+12.8.用數(shù)學(xué)歸納法證明“當(dāng)n為正偶數(shù)時(shí),xn-yn能被x+y整除”第一步應(yīng)驗(yàn)證n=時(shí),命題成立;第二步歸納假設(shè)成立應(yīng)寫(xiě)成. 答案2x2k-y2k能被x+y整除解析因?yàn)閚為正偶數(shù),故第一個(gè)值n=2,第二步假設(shè)n取第k個(gè)正偶數(shù)成立,即n=2k,故應(yīng)假設(shè)成x2k-y2k能被x+y整除.能力提升組9.用數(shù)學(xué)歸納法證明11×2+12×3+13×4+1n(n+1)=nn+1(nN*)時(shí),從“n=k”到“n=k+1”,等式左邊需增添的項(xiàng)是()A.1k(k+1)B.1k(k+1)+1(k+1)(k+2)C.1(k+1)(k+2)D.1k(k+2)答案C解析假設(shè)n=k時(shí),11×2+12×3+1k(k+1)=kk+1成立,那么n=k+1時(shí),11×2+12×3+1k(k+1)+1(k+1)(k+2)=kk+1+1(k+1)(k+2),所以從“kk+1”需增添的項(xiàng)是1(k+1)(k+2).故選C.10.利用數(shù)學(xué)歸納法證明“12n+1+12n+2+13n>13(n2,且nN*)”的過(guò)程中,由假設(shè)“n=k時(shí)”成立,推導(dǎo)“n=k+1時(shí)”也成立時(shí),該不等式左邊的變化是()A.增加13k+3B.增加13k+1+13k+2+13k+3C.增加13k+3并減少12k+1+12k+2D.增加13k+1+13k+2+13k+3并減少12k+1+12k+2答案D解析n=k+1時(shí),不等式為12k+3+12k+4+13k+3>13,增加13k+1+13k+2+13k+3并減少12k+1+12k+2.故選D.11.已知f(x)是定義域?yàn)檎麛?shù)集的函數(shù),對(duì)于定義域內(nèi)任意的k,若f(k)k2成立,則f(k+1)(k+1)2成立,下列命題成立的是()A.若f(3)9成立,且對(duì)于任意的k1,均有f(k)k2成立B.若f(4)16成立,則對(duì)于任意的k4,均有f(k)<k2成立C.若f(7)49成立,則對(duì)于任意的k<7,均有f(k)<k2成立D.若f(4)=25成立,則對(duì)于任意的k4,均有f(k)k2成立答案D解析因?yàn)閒(4)=25>42,所以對(duì)于k4,均有f(k)k2.僅有D選項(xiàng)符合題意.12.用數(shù)學(xué)歸納法證明3(2+7k)能被9整除,證明n=k+1時(shí),應(yīng)將3(2+7k+1)配湊成()A.6+21·7kB.3(2+7k)+21C.3(2+7k)D.21(2+7k)-36答案D解析要配湊出歸納假設(shè),即3(2+7k+1)=3(2+7·7k)=6+21·7k=21(2+7k)-36.故選D.13.設(shè)平面內(nèi)有n條直線(n3),其中有且僅有兩條直線互相平行,任意三條直線不過(guò)同一點(diǎn).若用f(n)表示這n條直線交點(diǎn)的個(gè)數(shù),則f(n)=()(n3).A.(n+1)(n-2)B.12(n+1)(n-2)C.n(n-1)D.12n(n-1)答案B解析f(3)=2,f(4)=f(3)+3=2+3=5,f(n)=f(3)+3+4+(n-1)=2+3+4+(n-1)=12(n+1)(n-2)(n3).14.若不等式1n+1+1n+2+13n+1>a24對(duì)一切正整數(shù)n都成立,正整數(shù)a的最大值為. 答案25解析當(dāng)n=1時(shí),11+1+11+2+13+1>a24,即2624>a24,所以a<26.而a是正整數(shù),所以取a=25,下面用數(shù)學(xué)歸納法證明1n+1+1n+2+13n+1>2524.(1)當(dāng)n=1時(shí),已證得不等式成立.(2)假設(shè)當(dāng)n=k(kN*)時(shí),不等式成立,即1k+1+1k+2+13k+1>2524.則當(dāng)n=k+1時(shí),有1(k+1)+1+1(k+1)+2+13(k+1)+1=1k+1+1k+2+13k+1+13k+2+13k+3+13k+4-1k+1>2524+13k+2+13k+4-23(k+1).因?yàn)?3k+2+13k+4-23(k+1)=6(k+1)(3k+2)(3k+4)-23(k+1)=18(k+1)2-2(9k2+18k+8)(3k+2)(3k+4)(3k+3)=2(3k+2)(3k+4)(3k+3)>0,所以當(dāng)n=k+1時(shí)不等式也成立.由(1)(2)知,對(duì)一切正整數(shù)n,都有1n+1+1n+2+13n+1>2524,所以a的最大值等于25.15.(2018浙江衢州模擬)在數(shù)列an中,已知a1=a(a>2),且an+1=an22(an-1)(nN*).(1)用數(shù)學(xué)歸納法證明an>2(nN*);(2)求證:an+1<an(nN*).證明(1)當(dāng)n=1時(shí),a1=a>2,命題成立.假設(shè)當(dāng)n=k(kN*,k1)時(shí),命題成立,即ak>2.則當(dāng)n=k+1時(shí),ak+1-2=ak22(ak-1)-2=(ak-2)22(ak-1)>0,所以當(dāng)n=k+1時(shí)ak+1>2也成立,由知對(duì)任意正整數(shù)n,都有an>2.(2)an+1-an=an22(an-1)-an=an(2-an)2(an-1),由(1)可知an>2>0,所以an+1<an.16.(2018浙江寧波效實(shí)中學(xué)高三期中)已知數(shù)列an,a1=3,an+1=3an-4an-1(nN*).(1)求a2,a3,a4的值,并猜想an的通項(xiàng)公式;(2)用數(shù)學(xué)歸納法證明你的猜想.(1)解因?yàn)閍1=3,且an+1=3an-4an-1,所以a2=3×3-43-1=52,a3=3×52-452-1=73,a1=3×73-473-1=94,由此猜想an=2n+1n.(2)證明當(dāng)n=1時(shí),a1=2×1+11=3,滿足要求,猜想成立;假設(shè)n=k(k1且kN*)時(shí),猜想成立,即ak=2k+1k,那么當(dāng)n=k+1時(shí),ak+1=3ak-4ak-1=3×2k+1k-42k+1k-1=2k+3k+1=2(k+1)+1k+1,這就表明當(dāng)n=k+1時(shí),猜想成立,根據(jù)可以斷定,對(duì)所有的正整數(shù)該猜想成立,即an=2n+1n.17.設(shè)a1=1,an+1=an2-2an+2+b(nN*).(1)若b=1,求a2,a3及數(shù)列an的通項(xiàng)公式.(2)若b=-1,問(wèn):是否存在實(shí)數(shù)c使得a2n<c<a2n+1對(duì)所有nN*成立?證明你的結(jié)論.(1)解法一a2=2,a3=2+1.再由題設(shè)條件知(an+1-1)2=(an-1)2+1.從而(an-1)2是首項(xiàng)為0,公差為1的等差數(shù)列,故(an-1)2=n-1,即an=n-1+1(nN*).解法二a2=2,a3=2+1.可寫(xiě)為a1=1-1+1,a2=2-1+1,a3=3-1+1.因此猜想an=n-1+1.下用數(shù)學(xué)歸納法證明上式:當(dāng)n=1時(shí)結(jié)論顯然成立.假設(shè)n=k時(shí)結(jié)論成立,即ak=k-1+1,則ak+1=(ak-1)2+1+1=(k-1)+1+1=(k+1)-1+1.這就是說(shuō),當(dāng)n=k+1時(shí)結(jié)論成立.所以an=n-1+1(nN*).(2)解法一設(shè)f(x)=(x-1)2+1-1,則an+1=f(an).令c=f(c),即c=(c-1)2+1-1,解得c=14.下用數(shù)學(xué)歸納法證明加強(qiáng)命題a2n<c<a2n+1<1.當(dāng)n=1時(shí),a2=f(1)=0,a3=f(0)=2-1,所以a2<14<a3<1,結(jié)論成立.假設(shè)n=k時(shí)結(jié)論成立,即a2k<c<a2k+1<1.易知f(x)在(-,1上為減函數(shù),從而c=f(c)>f(a2k+1)>f(1)=a2,即1>c>a2k+2>a2.再由f(x)在(-,1上為減函數(shù)得c=f(c)<f(a2k+2)<f(a2)=a3<1.故c<a2k+3<1,因此a2(k+1)<c<a2(k+1)+1<1.這就是說(shuō),當(dāng)n=k+1時(shí)結(jié)論成立.綜上,符合條件的c存在,其中一個(gè)值為c=14.解法二設(shè)f(x)=(x-1)2+1-1,則an+1=f(an).先證:0an1(nN*).當(dāng)n=1時(shí),結(jié)論明顯成立.假設(shè)n=k時(shí)結(jié)論成立,即0ak1.易知f(x)在(-,1上為減函數(shù),從而0=f(1)f(ak)f(0)=2-1<1.即0ak+11,這就是說(shuō),當(dāng)n=k+1時(shí)結(jié)論成立.故成立.再證:a2n<a2n+1(nN*).當(dāng)n=1時(shí),a2=f(1)=0,a3=f(a2)=f(0)=2-1,有a2<a3,即n=1時(shí)成立.假設(shè)n=k時(shí),結(jié)論成立,即a2k<a2k+1.由及f(x)在(-,1上為減函數(shù),得a2k+1=f(a2k)>f(a2k+1)=a2k+2,a2(k+1)=f(a2k+1)<f(a2k+2)=a2(k+1)+1.這就是說(shuō),當(dāng)n=k+1時(shí)成立.所以對(duì)一切nN*成立.由得a2n<a2n2-2a2n+2-1,即(a2n+1)2<a2n2-2a2n+2,因此a2n<14.又由,及f(x)在(-,1上為減函數(shù)得f(a2n)>f(a2n+1),即a2n+1>a2n+2.所以a2n+1>a2n+12-2a2n+1+2-1.解得a2n+1>14.綜上,由,知存在c=14使a2n<c<a2n+1對(duì)一切nN*成立.18.(2018浙江余姚中學(xué)模擬)已知數(shù)列an滿足a1=1,an+1=n2+n+1n2+nan+12n,nN*.(1)證明:當(dāng)n2時(shí),an2(nN*);(2)證明:an+1=11×2a1+12×3a2+1n×(n+1)an+2-12n(nN*).解(1)由題意,當(dāng)n=2時(shí),a2=32a1+12=22成立;當(dāng)n=k時(shí),假設(shè)ak2成立,則n=k+1時(shí)ak+1=k2+k+1k2+kak+12k=ak+1k2+kak+12k2+2k2+k+12k>2,所以n=k+1時(shí),ak+1>2成立.綜上可知,n2時(shí),an2.(2)由an+1=n2+n+1n2+nan+12n=an+1n(n+1)an+12n得an+1-an=1n(n+1)an+12n,所以a2-a1=11×2a1+121,a3-a2=12×3a2+122,a4-a3=13×4a3+123,an+1-an=1n(n+1)an+12n.所以an+1-a1=11×2a1+12×3a2+1n(n+1)+121+122+12n.又a1=1,所以an+1=11×2a1+12×3a2+1n(n+1)an+1+121-12n1-12=11×2a1+12×3a2+1n(n+1)an+2-12n.7