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課時(shí)規(guī)范練36數(shù)學(xué)歸納法基礎(chǔ)鞏固組1.如果命題p(n)對(duì)n=k(kN+)成立,則它對(duì)n=k+2也成立.若p(n)對(duì)n=2也成立,則下列結(jié)論正確的是()A.p(n)對(duì)所有正整數(shù)n都成立B.p(n)對(duì)所有正偶數(shù)n都成立C.p(n)對(duì)所有正奇數(shù)n都成立D.p(n)對(duì)所有自然數(shù)n都成立2.用數(shù)學(xué)歸納法證明命題“當(dāng)n是正奇數(shù)時(shí),xn+yn能被x+y整除”,在第二步時(shí),正確的證法是 ()A.假設(shè)n=k(kN+),證明n=k+1時(shí)命題成立B.假設(shè)n=k(k是正奇數(shù)),證明n=k+1時(shí)命題成立C.假設(shè)n=2k+1(kN+),證明n=k+1時(shí)命題成立D.假設(shè)n=k(k是正奇數(shù)),證明n=k+2時(shí)命題成立3.(2018安徽蚌埠期末,5)用數(shù)學(xué)歸納法證明不等式“1n+1+1n+2+12n1324(n2)”的過(guò)程中,歸納遞推由n=k到n=k+1時(shí),不等式的左邊()A.增加了一項(xiàng)12(k+1)B.增加了兩項(xiàng)12k+1+12(k+1)C.增加了兩項(xiàng)12k+1+12(k+1),又減少了一項(xiàng)1k+1D.增加了一項(xiàng)12(k+1),又減少了一項(xiàng)1k+14.(2018遼寧遼陽(yáng)期末,6)證明等式12+22+32+n2=n(n+1)(2n+1)6(nN+)時(shí),某學(xué)生的證明過(guò)程如下:(1)當(dāng)n=1時(shí),12=1236,等式成立;(2)假設(shè)n=k(kN+)時(shí),等式成立,即12+22+32+k2=k(k+1)(2k+1)6,則當(dāng)n=k+1時(shí),12+22+32+k2+(k+1)2=k(k+1)(2k+1)6+(k+1)2=(k+1)k(2k+1)+6(k+1)6=(k+1)(2k2+7k+6)6=(k+1)(k+1)+12(k+1)+16,所以當(dāng)n=k+1時(shí),等式也成立,故原等式成立.那么上述證明()A.全過(guò)程都正確B.當(dāng)n=1時(shí)驗(yàn)證不正確C.歸納假設(shè)不正確D.從n=k到n=k+1的推理不正確5.(2018遼寧撫順期中,14)用數(shù)學(xué)歸納法證明:“兩兩相交且不共點(diǎn)的n條直線把平面分為f(n)部分,則f(n)=1+n(n+1)2.”證明第二步歸納遞推時(shí),用到f(k+1)=f(k)+.6.試證:當(dāng)nN+時(shí),f(n)=32n+2-8n-9能被64整除.7.(2018山東師范大學(xué)附屬中學(xué)期中,18)證明:對(duì)任意的nN+,不等式3254762n+12nn+1成立.8.(2018廣東中山一中三模,21)設(shè)數(shù)列an滿足a1=3,an+1=an2-2nan+2(nN+).(1)求a2,a3,a4的值,并猜想數(shù)列an的通項(xiàng)公式(不需證明);(2)記Sn為數(shù)列an的前n項(xiàng)和,用數(shù)學(xué)歸納法證明:當(dāng)n6時(shí),有Sn2n成立.綜合提升組9.設(shè)f(x)是定義在正整數(shù)集上的函數(shù),且f(x)滿足:“當(dāng)f(k)k2成立時(shí),總可推出f(k+1)(k+1)2成立”.則下列命題總成立的是()A.若f(3)9成立,則當(dāng)k1時(shí),均有f(k)k2成立B.若f(5)25成立,則當(dāng)k5時(shí),均有f(k)k2成立C.若f(7)49成立,則當(dāng)k8時(shí),均有f(k)4時(shí),f(n)=(用n表示).11.(2018遼寧六校協(xié)作體期中,17)是否存在常數(shù)a,b使得等式12+22+n2=n(2n+1)(an+b)對(duì)一切正整數(shù)n都成立?若存在,求出a,b值,并用數(shù)學(xué)歸納法證明你的結(jié)論;若不存在,請(qǐng)說(shuō)明理由.創(chuàng)新應(yīng)用組12.(2018河南洛陽(yáng)模擬,18)將正整數(shù)作如下分組:(1),(2,3),(4,5,6),(7,8,9,10),(11,12,13,14,15),(16,17,18,19,20,21),.分別計(jì)算各組包含的正整數(shù)的和如下,S1=1,S2=2+3=5,S3=4+5+6=15,S4=7+8+9+10=34,S5=11+12+13+14+15=65,S6=16+17+18+19+20+21=111,(1)求S7的值;(2)由S1,S1+S3,S1+S3+S5,S1+S3+S5+S7的值,試猜測(cè)S1+S3+S2n-1的結(jié)果,并用數(shù)學(xué)歸納法證明.13.已知函數(shù)f0(x)=sinxx(x0),設(shè)fn(x)為fn-1(x)的導(dǎo)數(shù),nN+.(1)求2f12+2f22的值;(2)證明:對(duì)任意的nN+,等式nfn-14+4fn4=22都成立.參考答案課時(shí)規(guī)范練36數(shù)學(xué)歸納法1.Bn=k時(shí)成立,當(dāng)n=2時(shí),n=k+2成立,n為2,4,6,故n為所有正偶數(shù).2.D相鄰兩個(gè)正奇數(shù)相差2,故D選項(xiàng)正確.3.C當(dāng)n=k時(shí),左邊=1k+1+1k+2+12k,當(dāng)n=k+1時(shí),左邊=1k+2+1k+3+12k+1+12k+2,所以增加了兩項(xiàng)12k+1+12(k+1),又減少了一項(xiàng)1k+1,故答案為C.4.A考查所給的證明過(guò)程:當(dāng)n=1時(shí)驗(yàn)證是正確的,歸納假設(shè)是正確的,從n=k到n=k+1的推理也是正確的,即證明過(guò)程中不存在任何的問(wèn)題.故選A.5.k+1當(dāng)n=k(k2)時(shí),有f(k)=1+k(k+1)2,當(dāng)n=k+1時(shí),f(k+1)=1+(k+1)(k+2)2,從k到k+1左端需增加的代數(shù)式1+(k+1)(k+2)2-1-k(k+1)2=k+12(k+2-k)=k+1,在證明第二步歸納推理的過(guò)程中,用到f(k+1)=f(k)+(k+1).6.證明 (1)當(dāng)n=1時(shí),f(1)=64,命題顯然成立.(2)假設(shè)當(dāng)n=k(kN+,k1)時(shí),f(k)=32k+2-8k-9能被64整除,則當(dāng)n=k+1時(shí),f(k+1)=32(k+1)+2-8(k+1)-9=9(32k+2-8k-9)+98k+99-8(k+1)-9=9(32k+2-8k-9)+64(k+1),即f(k+1)=9f(k)+64(k+1),因此當(dāng)n=k+1時(shí)命題也成立.根據(jù)(1)(2)可知,對(duì)于任意nN+,命題都成立.7.證明 當(dāng)n=1時(shí),左邊=,右邊=2,因?yàn)?,所以不等式成立.假設(shè)當(dāng)n=k時(shí)不等式成立,即3254762k+12kk+1成立.則當(dāng)n=k+1時(shí),左邊3254762k+12k2k+32k+2k+12k+32k+2=(2k+3)24(k+1)=4(k+1)2+4(k+1)+14(k+1)=(k+1)+1+14(k+1)(k+1)+1,所以當(dāng)n=k+1時(shí),不等式也成立.由可得不等式恒成立.8.解 (1)a2=5,a3=7,a4=9,猜想an=2n+1.(2)Sn=n(3+2n+1)2=n2+2n,下證:n6(nN+)時(shí)都有2nn2+2n.當(dāng)n=6時(shí),2662+26,即6448成立;假設(shè)n=k(k6,kN+)時(shí),2kk2+2k成立,那么當(dāng)n=k+1時(shí),2k+1=22k2(k2+2k)=k2+2k+k2+2kk2+2k+3+2k=(k+1)2+2(k+1),即n=k+1時(shí),不等式成立.故對(duì)于所有的n6(nN+),都有2nn2+2n成立.9.D對(duì)A,當(dāng)k=1或2時(shí),不一定有f(k)k2成立;對(duì)B,只能得出:對(duì)于任意的k5,均有f(k)k2成立,不能得出:對(duì)任意的k5,均有f(k)k2成立;對(duì)C,若f(7)49成立不能推出任何結(jié)論;對(duì)D,f(4)=2516,對(duì)于任意的k4,均有f(k)k2成立.故選D.10.5 (n+1)(n-2)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).11.解 分別令n=1,2,可得1=3(a+b),5=10(2a+b),解得a=16,b=16.故猜想等式12+22+n2=n(2n+1)(n+1)6對(duì)一切正整數(shù)n都成立.下面用數(shù)學(xué)歸納法證明:當(dāng)n=1時(shí),由上面的探求可知等式成立.假設(shè)n=k(kN+,k1)時(shí)猜想成立,即12+22+k2=k(2k+1)(k+1)6.當(dāng)n=k+1時(shí),12+22+k2+(k+1)2=k(2k+1)(k+1)6+(k+1)2=(k+1)k(2k+1)+6(k+1)6=(k+1)(2k2+7k+6)6=(k+1)(2k+3)(k+2)6.所以當(dāng)n=k+1時(shí),等式也成立.由知猜想成立,即存在a=16,b=16使命題成立.12.解 (1)S7=22+23+24+25+26+27+28=175.(2)S1=1;S1+S3=16;S1+S3+S5=81;S1+S3+S5+S7=256;猜測(cè)S1+S3+S5+S2n-1=n4.證明如下:記Mn=S1+S3+S5+S2n-1,當(dāng)n=1時(shí),猜想成立.設(shè)當(dāng)n=k時(shí),命題成立,即Mk=S1+S3+S5+S2k-1=k4.下面證明當(dāng)n=k+1時(shí),猜想也成立.事實(shí)上,由題設(shè)可知Sn是由1+2+3+(n-1)+1=n(n-1)2+1開(kāi)始的n個(gè)連續(xù)自然數(shù)的和.所以Sn=n(n-1)2+1+n(n-1)2+2+n(n-1)2+n=n(n2+1)2,所以S2k+1=(2k+1)(2k+1)2+12=(2k+1)(2k2+2k+1)=4k3+6k2+4k+1,從而Mk+1=Mk+S2k+1=k4+4k3+6k2+4k+1=(k+1)4,所以猜想在n=k+1時(shí)也成立.綜合(1)(2)可知猜想對(duì)任何nN+都成立.13.(1)解 由已知,得f1(x)=f0(x)=sinxx=cosxx-sinxx2,于是f2(x)=f1(x)=cosxx-sinxx2=-sinxx-2cosxx2+2sinxx3,所以f12=-42,f22=-2+163,故2f12+2f22=-1.(2)證明 由已知,得xf0(x)=sin x,等式兩邊分別對(duì)x求導(dǎo),得f0(x)+xf0(x)=cos x,即f0(x)+xf1(x)=cos x=sinx+,類(lèi)似可得,2f1(x)+xf2(x)=-sin x=sin(x+),3f2(x)+xf3(x)=-cos x=sinx+32,4f3(x)+xf4(x)=sin x=sin(x+2).下面用數(shù)學(xué)歸納法證明等式nfn-1(x)+xfn(x)=sinx+n2對(duì)所有的xN+都成立.當(dāng)n=1時(shí),由上可知等式成立.假設(shè)當(dāng)n=k時(shí),等式成立,即kfk-1(x)+xfk(x)=sinx+k2.因?yàn)閗fk-1(x)+xfk(x)=kfk-1(x)+fk(x)+xfk(x)=(k+1)fk(x)+xfk+1(x),sinx+k2=cosx+k2x+k2=sinx+(k+1)2,所以(k+1)fk(x)+xfk+1(x)=sinx+(k+1)2.因此當(dāng)n=k+1時(shí),等式也成立.綜合可知等式nfn-1(x)+xfn(x)=sinx+n2對(duì)所有的nN+都成立.令x=4,可得nfn-14+4fn4=sin 4+n2(nN+),所以nfn-14+4fn4=22(nN+).