2023-2024學(xué)年高中數(shù)學(xué)人教A版2019選擇性課后習(xí)題第四章　數(shù)列4-4　數(shù)學(xué)歸納法

4.4*數(shù)學(xué)歸納法必備知識基礎(chǔ)練1.利用數(shù)學(xué)歸納法證明不等式1+12+13+…+12n-1<n(n≥2,n∈N*)的過程中,由n=k變到A.1項(xiàng) B.k項(xiàng) C.2k1項(xiàng) D.2k項(xiàng)2.(2021上海交大附中高一下入學(xué)檢測)用數(shù)學(xué)歸納法證明“對任意偶數(shù)n,anbn能被ab整除”時(shí),其第二步論證應(yīng)該是()A.假設(shè)n=k(k為正整數(shù))時(shí)命題成立,再證n=k+1時(shí)命題也成立B.假設(shè)n=2k(k為正整數(shù))時(shí)命題成立,再證n=2k+1時(shí)命題也成立C.假設(shè)n=k(k為正整數(shù))時(shí)命題成立,再證n=2k+1時(shí)命題也成立D.假設(shè)n=2k(k為正整數(shù))時(shí)命題成立,再證n=2(k+1)時(shí)命題也成立3.(2021上海黃浦高二期末)用數(shù)學(xué)歸納法證明:(n+1)(n+2)…(n+n)=2n×1×3×…×(2n1)(n∈N*).從n=k(k∈N*)到n=k+1,若設(shè)f(k)=(k+1)(k+2)…(k+k),則f(k+1)=()A.f(k)+[2(2k+1)]B.f(k)·[2(2k+1)]C.f(k)+2D.f(k)·24.(多選題)對于不等式n2+n≤n+1(n∈N*①當(dāng)n=1時(shí),12+1≤1+1,②假設(shè)n=k(k∈N*)時(shí),不等式成立,即k2+k<k+1,則n=k+(k(k2+3k+2)∴當(dāng)n=k+1時(shí),不等式成立,關(guān)于上述證明過程的說法正確的是()A.證明過程全都正確B.當(dāng)n=1時(shí)的驗(yàn)證正確C.歸納假設(shè)正確D.從n=k到n=k+1的推理不正確5.(多選題)一個(gè)與正整數(shù)n有關(guān)的命題,當(dāng)n=2時(shí)命題成立,且由n=k時(shí)命題成立可以推得n=k+2時(shí)命題也成立,則下列說法正確的是()A.該命題對于n=6時(shí)命題成立B.該命題對于所有的正偶數(shù)都成立C.該命題何時(shí)成立與k取值無關(guān)D.以上答案都不對6.用數(shù)學(xué)歸納法證明112+13-14+…+12n-1-12n=1n+1+1n7.用數(shù)學(xué)歸納法證明:1222+3242+…+(1)n1n2=(1)n1·n(n+1)2(n8.(2021陜西西安鐵路一中高二期末)在數(shù)列{an}中,a1=12,an+1=3(1)求出a2,a3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.關(guān)鍵能力提升練9.(2021江西贛州高二期末)用數(shù)學(xué)歸納法證明不等式1n+1+1n+2+…+1n+n>1314(n∈N*)的過程中A.增加了12B.增加了12C.增加了12D.增加了110.(2021浙江溫州期中)利用數(shù)學(xué)歸納法證明等式:1·n+2·(n1)+3·(n2)+…+n·1=16n(n+1)(n+2)(n∈N*),當(dāng)n=k時(shí),左邊的和1·k+2·(k1)+3·(k2)+…+k·1,記作Sk,則當(dāng)n=k+1時(shí)左邊的和,記作Sk+1,則Sk+1Sk=(A.1+2+3+…+kB.1+2+3+…+(k1)C.1+2+3+…+(k+1)D.1+2+3+…+(k2)11.(多選題)用數(shù)學(xué)歸納法證明2n-12n+1>nn+1對任意n≥λ(n,λ∈NA.1 B.2 C.3 D.412.記凸k邊形的內(nèi)角和為f(k),則凸k+1邊形的內(nèi)角和f(k+1)=f(k)+.

13.是否存在a,b,c使等式1n2+2n2+3n2+…+nn2=an14.用數(shù)學(xué)歸納法證明:23×45×67×…×215.已知數(shù)列{fn(x)}滿足f1(x)=x1+x2(x>0),fn+1(x)=f1(fn((1)求f2(x),f3(x),并猜想{fn(x)}的通項(xiàng)公式;(2)用數(shù)學(xué)歸納法證明猜想.學(xué)科素養(yǎng)創(chuàng)新練16.已知數(shù)列{an}滿足a1=2,an+1=an2nan+1(n∈N*(1)求a2,a3,a4,并由此猜想出{an}的一個(gè)通項(xiàng)公式并用數(shù)學(xué)歸納法證明;(2)用數(shù)學(xué)歸納法證明:當(dāng)n>1時(shí),1a1+1a參考答案4.4*數(shù)學(xué)歸納法1.D當(dāng)n=k時(shí),不等式左邊的最后一項(xiàng)為12k-1,而當(dāng)n=k+1時(shí),最后一項(xiàng)為12k+1-2.D根據(jù)證明的結(jié)論,n為正偶數(shù),故第二步的假設(shè)應(yīng)寫成:假設(shè)n=2k,k∈N*時(shí)命題正確,即當(dāng)n=2k,k∈N*時(shí),a2kb2k能被ab整除,再推證n=2k+2時(shí)正確.故選D.3.B由數(shù)學(xué)歸納法證明(n+1)(n+2)…(n+n)=2n×1×3×…×(2n1)(n∈N*)時(shí),從“k”到“k+1”的證明,左邊需增添的一個(gè)因式是(2k+1)(2k+2)k+1=2(2k+1),則f(k+4.BCDn=1的驗(yàn)證及歸納假設(shè)都正確,但從n=k到n=k+1的推理中沒有使用歸納假設(shè),而通過不等式的放縮法直接證明,不符合數(shù)學(xué)歸納法的證題要求.故選BCD.5.AB由n=k時(shí)命題成立可以推出n=k+2時(shí)命題也成立,且n=2時(shí),命題成立,故對所有的正偶數(shù)都成立.故選AB.6.112=1212k+1-12(k+1)當(dāng)n=1時(shí),7.證明①當(dāng)n=1時(shí),左邊=12=1,右邊=(1)0×1×(左邊=右邊,等式成立.②假設(shè)n=k(k∈N*)時(shí),等式成立,即1222+3242+…+(1)k1k2=(1)k1·k(則當(dāng)n=k+1時(shí),1222+3242+…+(1)k1k2+(1)k(k+1)2=(1)k1·k(k+1)2+(1)=(1)k(k+1)·(k+1)k2=(1)k·(k∴當(dāng)n=k+1時(shí),等式也成立,根據(jù)①②可知,對于任何n∈N*等式成立.8.(1)解由a1=12,an+1=3anan+3,得a2=3猜想an=3n(2)證明①當(dāng)n=1時(shí),a1=12=36=②假設(shè)當(dāng)n=k(k∈N*)時(shí),結(jié)論成立,即ak=3k那么,當(dāng)n=k+1時(shí),ak+1=3aka由①和②可知對任意n∈N*,都有an=3n+59.D當(dāng)n=k時(shí),1k+1+1k當(dāng)n=k+1時(shí),1k+1+1+1k左邊增加了1210.C依題意,Sk=1·k+2·(k1)+3·(k2)+…+k·1,則Sk+1=1·(k+1)+2·k+3·(k1)+4·(k2)+…+k·2+(k+1)·1,∴Sk+1Sk=1·[(k+1)k]+2·[k(k1)]+3·[(k1)(k2)]+4·[(k2)(k3)]+…+k·(21)+(k+1)·1=1+2+3+…+k+(k+1).11.CD取n=1,則2n-取n=2,則2n-取n=3,則2n-取n=4,則2n-猜想當(dāng)n≥3時(shí),2n-12n+1>n證明:當(dāng)n=3時(shí),2n-設(shè)當(dāng)n=k(k≥3,k∈N*)時(shí),有2k-則當(dāng)n=k+1時(shí),有2k令t=2k-12k+1,因?yàn)閠>kk+1,故2k+1因?yàn)?k+14k+3所以當(dāng)n=k+1時(shí),不等式也成立,由數(shù)學(xué)歸納法可知2n-12n+1>nn12.π由凸k邊形變?yōu)橥筴+1邊形時(shí),增加了一個(gè)三角形圖形,故f(k+1)=f(k)+π.13.解取n=1,2,3可得a解得a=13,b=12,c=下面用數(shù)學(xué)歸納法證明1n2+2n即證12+22+…+n2=16n(n+1)(2n+①當(dāng)n=1時(shí),左邊=1,右邊=1,∴等式成立;②假設(shè)當(dāng)n=k(k∈N*)時(shí)等式成立,即12+22+…+k2=16k(k+1)(2k+1)成立則當(dāng)n=k+1時(shí),等式左邊=12+22+…+k2+(k+1)2=16k(k+1)(2k+1)+(k+1)=16[k(k+1)(2k+1)+6(k+1)2=16(k+1)(2k2+7k+6)=16(k+1)(k+2)(2故當(dāng)n=k+1時(shí)等式成立.由數(shù)學(xué)歸納法,綜合①②當(dāng)n∈N*等式成立,故存在a=13,b=12,c=114.證明①∵當(dāng)n=1時(shí),49-12∴49∴23<12=11+1,②假設(shè)當(dāng)n=k(k∈N*)時(shí),不等式成立,即23×45×則當(dāng)n=k+1時(shí),23×45×∵2k+12k+321(4(k∴2k+12k+32<1(∴2k+12k+3<1(綜合①②可知,對于任意n∈N*,23×45×615.解(1)f2(x)=f1[f1(x)]=f1(x)1+f12(x)=x1+2x2,f猜想:fn(x)=x1+nx2(n∈(2)下面用數(shù)學(xué)歸納法證明fn(x)=x1+nx2(n①當(dāng)n=1時(shí),f1(x)=x1+x2②假設(shè)當(dāng)n=k(k∈N*)時(shí),猜想成立,即fk(x)=x1+則當(dāng)n=k+1時(shí),fk+1=f1[fk(x)]=x1+kx21+x1+k結(jié)合①②可知,猜想fn(x)=x1+nx2對一切n∈16.(1)解由a1=2,得a2=a12a1+1由a2=3,得a3=a222a2+1由a3=4,得a4=a323a3+1由此猜想an的一個(gè)通項(xiàng)公式為an=n+1.下面證明an=n+1.當(dāng)n=1時(shí),a2=2=1+1,成立.假設(shè)當(dāng)n=k(k≥2)