2024-2025學(xué)年高中數(shù)學(xué)第四講數(shù)學(xué)歸納法證明不等式測(cè)評(píng)練習(xí)含解析新人教A版選修4-5

PAGEPAGE1第四講測(cè)評(píng)(時(shí)間:120分鐘滿分:150分)一、選擇題(本大題共12小題,每小題5分,共60分)1.用數(shù)學(xué)歸納法證明3n≥n3(n≥3,n∈N)第一步應(yīng)驗(yàn)證()A.n=1 B.n=2 C.n=3 D.n=4解析由n≥3,n∈N知,應(yīng)驗(yàn)證n=3.答案C2.在用數(shù)學(xué)歸納法證明等式1+2+3+…+2n=2n2+n(n∈N+)的第(2)步中,假設(shè)當(dāng)n=k時(shí)原等式成立,則在n=k+1時(shí)須要證明的等式為()A.1+2+3+…+2k+2(k+1)=2k2+k+2(k+1)2+(k+1)B.1+2+3+…+2k+2(k+1)=2(k+1)2+(k+1)C.1+2+3+…+2k+2k+1+2(k+1)=2k2+k+2(k+1)2+(k+1)D.1+2+3+…+2k+2k+1+2(k+1)=2(k+1)2+(k+1)解析用數(shù)學(xué)歸納法證明等式1+2+3+…+2n=2n2+n時(shí),當(dāng)n=1時(shí)左邊所得的項(xiàng)是1+2=3,右邊=2×12+1=3,命題成立.假設(shè)當(dāng)n=k時(shí)命題成立,即1+2+3+…+2k=2k2+k.則當(dāng)n=k+1時(shí),左邊為1+2+3+…+2k+2k+1+2(k+1),故從“k→k+1”需增加的項(xiàng)是2k+1+2(k+1),因此1+2+3+…+2k+2k+1+2(k+1)=2(k+1)2+(k+1).答案D3.記等式1·n+2·(n-1)+3·(n-2)+…+n·1=n(n+1)(n+2)左邊的式子為f(n),用數(shù)學(xué)歸納法證明該等式的其次步歸納遞推時(shí),即當(dāng)n從k變?yōu)閗+1時(shí),等式左邊的變更量f(k+1)-f(k)=()A.k+1 B.1·(k+1)+(k+1)·1C.1+2+3+…+k D.1+2+3+…+k+(k+1)解析依題意,f(k)=1·k+2·(k-1)+3·(k-2)+…+k·1,則f(k+1)=1·(k+1)+2·k+3·(k-1)+4·(k-2)+…+k·2+(k+1)·1,∴f(k+1)-f(k)=1·[(k+1)-k]+2·[k-(k-1)]+3·[(k-1)-(k-2)]+4·[(k-2)-(k-3)]+…+k·(2-1)+(k+1)·1=1+2+3+…+k+(k+1).答案D4.用數(shù)學(xué)歸納法證明“n3+(n+1)3+(n+2)3(n∈N+)能被9整除”,要利用歸納假設(shè)證當(dāng)n=k+1時(shí)的狀況,只需綻開(kāi)()A.(k+3)3 B.(k+2)3C.(k+1)3 D.(k+1)3+(k+2)3解析當(dāng)n=k+1時(shí),證明“(k+1)3+(k+2)3+(k+3)3能被9整除”,依據(jù)歸納假設(shè),當(dāng)n=k時(shí),證明“k3+(k+1)3+(k+2)3能被9整除”,所以只需綻開(kāi)(k+3)3.答案A5.用數(shù)學(xué)歸納法證明2n≥n2(n≥5,n∈N+)成立時(shí),其次步歸納假設(shè)的正確寫(xiě)法是()A.假設(shè)當(dāng)n=k時(shí)命題成立B.假設(shè)當(dāng)n=k(k∈N+)時(shí)命題成立C.假設(shè)當(dāng)n=k(k≥5)時(shí)命題成立D.假設(shè)當(dāng)n=k(k>5)時(shí)命題成立解析由數(shù)學(xué)歸納法的步驟可知,選項(xiàng)C正確.答案C6.用數(shù)學(xué)歸納法證明“Sn=+…+>1(n∈N+)”時(shí),S1等于()A. B.C. D.解析當(dāng)n=1時(shí),S1=.答案D7.已知在數(shù)列{an}中,a1=1,a2=2,an+1=2an+an-1(n∈N+),用數(shù)學(xué)歸納法證明a4n能被4整除,假設(shè)a4k能被4整除,然后應(yīng)當(dāng)證明()A.a4k+1能被4整除 B.a4k+2能被4整除C.a4k+3能被4整除 D.a4k+4能被4整除解析由假設(shè)a4k能被4整除,則當(dāng)n=k+1時(shí),應(yīng)當(dāng)證明a4(k+1)=a4k+4能被4整除.答案D8.設(shè)0<θ<,已知a1=2cosθ,an+1=,則猜想an為()A.2cos B.2cosC.2cos D.2sin解析a1=2cosθ,a2==2cos,a3==2cos,猜想an=2cos.答案B9.從一樓到二樓的樓梯共有n級(jí)臺(tái)階,每步只能跨上1級(jí)或2級(jí),走完這n級(jí)臺(tái)階共有f(n)種走法,則下面的猜想正確的是()A.f(n)=f(n-1)+f(n-2)(n≥3)B.f(n)=2f(n-1)(n≥2)C.f(n)=2f(n-1)-1(n≥2)D.f(n)=f(n-1)f(n-2)(n≥3)解析分別取n=1,2,3,4驗(yàn)證,得f(n)=答案A10.用數(shù)學(xué)歸納法證明“34n+1+52n+1(n∈N+)能被8整除”時(shí),若當(dāng)n=k時(shí)命題成立,欲證當(dāng)n=k+1時(shí)命題成立,對(duì)于34(k+1)+1+52(k+1)+1可變形為()A.56×34k+1+25(34k+1+52k+1)B.34×34k+1+52×52kC.34k+1+52k+1D.25(34k+1+52k+1)解析由于34(k+1)+1+52(k+1)+1=81×34k+1+25×52k+1+25×34k+1-25×34k+1=56×34k+1+25(34k+1+52k+1),故應(yīng)選A.答案A11.下列說(shuō)法正確的是()A.若一個(gè)命題當(dāng)n=1,2時(shí)為真,則此命題為真命題B.若一個(gè)命題當(dāng)n=k時(shí)成立且推得n=k+1時(shí)也成立,則這個(gè)命題為真命題C.若一個(gè)命題當(dāng)n=1,2時(shí)為真,則當(dāng)n=3時(shí)這個(gè)命題也為真D.若一個(gè)命題當(dāng)n=1時(shí)為真,n=k時(shí)為真能推得n=k+1時(shí)亦為真,則此命題為真命題解析由數(shù)學(xué)歸納法可知,只有當(dāng)n的初始取值成立且由n=k成立能推得n=k+1時(shí)也成立時(shí),才可以證明結(jié)論正確,二者缺一不行.A,B,C項(xiàng)均不全面.答案D12.若命題A(n)(n∈N+)在n=k(k∈N+)時(shí)成立,則有當(dāng)n=k+1時(shí)命題也成立.現(xiàn)知命題對(duì)n=n0(n0∈N+)時(shí)成立,則有()A.命題對(duì)全部正整數(shù)都成立B.命題對(duì)小于n0的正整數(shù)不成立,對(duì)大于或等于n0的正整數(shù)都成立C.命題對(duì)小于n0的正整數(shù)成立與否不能確定,對(duì)大于或等于n0的正整數(shù)都成立D.以上說(shuō)法都不正確解析數(shù)學(xué)歸納法證明的結(jié)論只是對(duì)n的初始值及后面的正整數(shù)成立,而對(duì)于初始值前的正整數(shù)不肯定成立.答案C二、填空題(本大題共4小題,每小題5分,共20分)13.用數(shù)學(xué)歸納法證明cosα+cos3α+…+cos(2n-1)α=(sinα≠0,n∈N),在驗(yàn)證n=1時(shí),等式右邊的式子是.

解析當(dāng)n=1時(shí),右邊==cosα.答案cosα14.設(shè)f(n)=,用數(shù)學(xué)歸納法證明f(n)≥3.在“假設(shè)當(dāng)n=k時(shí)成立”后,f(k+1)與f(k)的關(guān)系是f(k+1)=f(k)·.

解析當(dāng)n=k時(shí),f(k)=,當(dāng)n=k+1時(shí),f(k+1)=,所以f(k)應(yīng)乘.答案15.用數(shù)學(xué)歸納法證明+…+,假設(shè)當(dāng)n=k時(shí),不等式成立,則當(dāng)n=k+1時(shí),應(yīng)推證的目標(biāo)是.

解析留意不等式兩邊含變量“n”的式子,因此當(dāng)n=k+1時(shí),應(yīng)當(dāng)是含“n”的式子發(fā)生變更,所以當(dāng)n=k+1時(shí),應(yīng)為+…+.答案+…+16.導(dǎo)學(xué)號(hào)26394070設(shè)a,b均為正實(shí)數(shù),n∈N+,已知M=(a+b)n,N=an+nan-1b,則M,N的大小關(guān)系為.

解析由貝努利不等式(1+x)n>1+nx(x>-1,且x≠0,n>1,n∈N+),可知當(dāng)n>1時(shí),令x=,所以>1+n·,所以>1+n·,即(a+b)n>an+nan-1b.當(dāng)n=1時(shí),M=N,故M≥N.答案M≥N三、解答題(本大題共6小題,共70分)17.(本小題滿分10分)用數(shù)學(xué)歸納法證明:12-22+32-42+…+(2n-1)2-(2n)2=-n(2n+1)(n∈N+).證明(1)當(dāng)n=1時(shí),左邊=12-22=-3,右邊=-1×(2×1+1)=-3,等式成立.(2)假設(shè)當(dāng)n=k(k∈N+,k≥1)時(shí)等式成立,即12-22+32-42+…+(2k-1)2-(2k)2=-k(2k+1).當(dāng)n=k+1時(shí),12-22+32-42+…+(2k-1)2-(2k)2+(2k+1)2-[2(k+1)]2=-k(2k+1)+(2k+1)2-[2(k+1)]2=-2k2-5k-3=-(k+1)(2k+3)=-(k+1)[2(k+1)+1],即當(dāng)n=k+1時(shí),等式成立.由(1)(2)可知,對(duì)任何n∈N+,等式成立.18.(本小題滿分12分)求證:兩個(gè)連續(xù)正整數(shù)的積能被2整除.證明設(shè)n∈N+,則要證明n(n+1)能被2整除.(1)當(dāng)n=1時(shí),1×(1+1)=2,能被2整除,即命題成立.(2)假設(shè)n=k(k≥1)時(shí)命題成立,即k·(k+1)能被2整除.當(dāng)n=k+1時(shí),(k+1)(k+1+1)=(k+1)(k+2)=k(k+1)+2(k+1),由歸納假設(shè)k(k+1)及2(k+1)都能被2整除,所以(k+1)(k+2)能被2整除,故當(dāng)n=k+1時(shí)命題成立.由(1)(2)可知,命題對(duì)一切n∈N+都成立.19.(本小題滿分12分)設(shè)函數(shù)fn(x)=x+x2+…+xn-2(n∈N,n≥2),當(dāng)x>-1,且x≠0時(shí),證明:fn(x)>0恒成立.(x+1)n=x0+x+x2+…+xn,,m,n∈N+,且n≥m證明要證fn(x)>0恒成立,因?yàn)閤>-1,且x≠0,所以只需證·x+·x2+…+xn>1+nx,即證(1+x)n>1+nx.(1)當(dāng)n=2時(shí),不等式成立.(2)假設(shè)當(dāng)n=k(k≥2)時(shí)不等式成立,即(1+x)k>1+kx.當(dāng)n=k+1時(shí),有(1+x)k+1=(1+x)k·(1+x)>(1+kx)(1+x)=1+(k+1)x+kx2>1+(k+1)x,即當(dāng)n=k+1時(shí)不等式成立.由(1)(2)可知,對(duì)隨意n∈N,n≥2,(1+x)n>1+nx成立,即fn(x)>0恒成立.20.(本小題滿分12分)已知點(diǎn)的序列An(xn,0),n∈N+,其中x1=0,x2=a(a>0),A3是線段A1A2的中點(diǎn),A4是線段A2A3的中點(diǎn),…,An是線段An-2An-1的中點(diǎn),….(1)寫(xiě)出xn與xn-1,xn-2之間的關(guān)系式(n≥3);(2)設(shè)an=xn+1-xn,計(jì)算a1,a2,a3,由此推想數(shù)列{an}的通項(xiàng)公式,并加以證明.解(1)當(dāng)n≥3時(shí),xn=.(2)a1=x2-x1=a,a2=x3-x2=-x2=-(x2-x1)=-a,a3=x4-x3=-x3=-(x3-x2)=-a.由此推想an=a(n∈N+).用數(shù)學(xué)歸納法證明:①當(dāng)n=1時(shí),a1=x2-x1=a=a,通項(xiàng)公式成立.②假設(shè)當(dāng)n=k時(shí),ak=a成立.當(dāng)n=k+1時(shí),ak+1=xk+2-xk+1=-xk+1=-(xk+1-xk)=-ak=-a=a,通項(xiàng)公式成立.由①②知,an=a(n∈N+)成立.21.導(dǎo)學(xué)號(hào)26394071(本小題滿分12分)求證:tanα·tan2α+tan2α·tan3α+…+tan(n-1)α·tannα=-n(n≥2,n∈N+).證明(1)當(dāng)n=2時(shí),左邊=tanα·tan2α,右邊=-2=-2=-2==tanα·tan2α=左邊,等式成立.(2)假設(shè)當(dāng)n=k(k≥2)時(shí)等式成立,即tanα·tan2α+tan2α·tan3α+…+tan(k-1)α·tankα=-k.當(dāng)n=k+1時(shí),tanα·tan2α+tan2α·tan3α+…+tan(k-1)α·tankα+tankα·tan(k+1)α=-k+tankα·tan(k+1)α=-k=[1+tan(k+1)α·tanα]-k=[tan(k+1)α-tanα]-k=-(k+1),所以當(dāng)n=k+1時(shí)等式成立.由(1)和(2)知,當(dāng)n≥