2025屆高三一輪復(fù)習(xí)數(shù)學(xué)試題(人教版新高考新教材)考點(diǎn)規(guī)范練29　數(shù)學(xué)歸納法

考點(diǎn)規(guī)范練29數(shù)學(xué)歸納法一、基礎(chǔ)鞏固1.對(duì)于不等式n2+n<n+1(n∈(1)當(dāng)n=1時(shí),12+1<1+(2)假設(shè)當(dāng)n=k(k∈N*)時(shí),不等式成立,即k2+k<k+1,則當(dāng)n=k+1時(shí),(k+1)2故當(dāng)n=k+1時(shí),不等式成立.在上述證明中,()A.過(guò)程全部正確B.對(duì)于當(dāng)n=1時(shí)結(jié)論的推理不正確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.2.用數(shù)學(xué)歸納法證明:1+12+13+14+…+12n證明:(1)當(dāng)n=1時(shí),左邊=1,右邊=1,不等式成立.(2)假設(shè)當(dāng)n=k(k∈N*)時(shí),不等式成立,即有1+12+13+14那么當(dāng)n=k+1時(shí),左邊=1+12+13+14+…+12k-1+又12k+12k+1+…+12k+1-1<12k×2k=即當(dāng)n=k+1時(shí),不等式也成立.由(1)(2)可知,對(duì)于任意n∈N*,1+12+13+14+3.已知數(shù)列{xn},{yn}滿足x1=5,y1=-5,2xn+1+3yn=7,6xn+yn+1=13.求證:xn=3n+2,yn=1-2×3n(n∈N*).證明:(1)當(dāng)n=1時(shí),x1=5,31+2=5,y1=-5,且1-2×31=-5,即等式成立.(2)假設(shè)當(dāng)n=k(k∈N*)時(shí),等式成立,即xk=3k+2,yk=1-2×3k,那么當(dāng)n=k+1時(shí),由2xk+1+3yk=7,得xk+1=12(7-3yk)=7-3(1-2由6xk+yk+1=13,得yk+1=13-6xk=13-6(3k+2)=1-2×3k+1;故當(dāng)n=k+1時(shí),等式也成立.由(1)(2)可知,xn=3n+2,yn=1-2×3n對(duì)一切n∈N*都成立.二、綜合應(yīng)用4.設(shè)平面內(nèi)有n(n∈N*,n≥3)條直線,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(4)=;當(dāng)n>4時(shí),f(n)=(用n表示).

答案:512(n+1)(n-解析:由題意知f(3)=2,f(4)=5,f(5)=9,可以歸納出每增加一條直線,交點(diǎn)增加的個(gè)數(shù)為原有直線的條數(shù),即f(4)-f(3)=3,f(5)-f(4)=4,猜測(cè)得出f(n)-f(n-1)=n-1(n≥4),則f(n)-f(3)=3+4+…+(n-1),故f(n)=12(n+1)(n-2)5.用數(shù)學(xué)歸納法證明:1-12+13?14+…+12n-1證明:(1)當(dāng)n=1時(shí),左邊=1-12=12,右邊=(2)假設(shè)當(dāng)n=k(k∈N*)時(shí),等式成立,即1-12+13?14+…那么當(dāng)n=k+1時(shí),1-12+13=1k+1+1=1k+2+…=1k+2+1=1(k+1)+1故當(dāng)n=k+1時(shí),等式也成立.根據(jù)(1)(2)可知,等式對(duì)于任何n∈N*都成立.6.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,a1=2,Sn+1=2Sn+2,數(shù)列{bn}滿足b1=1,bn+1=bn2+bn,其中n∈N(1)證明:數(shù)列{an}是等比數(shù)列;(2)記Tn=b12+b22+…+bn2,證明:S證明:(1)已知Sn+1=2Sn+2,當(dāng)n≥2時(shí),Sn=2Sn-1+2,兩式相減,得an+1=2an(n≥2).又S2=2S1+2=2a1+2=6,即a1+a2=6,所以a2=6-a1=4,即a2a故數(shù)列{an}是首項(xiàng)為2,公比為2的等比數(shù)列.(2)Sn=2(1-2n)要證明Sn-2Tn≤2,即證明2n+1-2-2Tn≤2,即Tn≥2n-2,①當(dāng)n=1時(shí),T1=b12=1,21-2=0,此時(shí)T1>21-2②假設(shè)n=k時(shí),不等式成立,即Tk≥2k-2,那么當(dāng)n=k+1時(shí),Tk+1=b12+b22+…+bk由bn+1=bn2+bn,知bn+1-bn=bn2>0,所以bn+1>bn≥b由bn+1=bn2+bn,知bn+1bn=b所以bk+1=b1·b2b1·…·bk所以Tk+1=b12+b22+…+bk2+bk+12≥2綜上①②可知,不等式Sn-2Tn≤2對(duì)任何n∈N*都成立.三、探究創(chuàng)新7.設(shè)數(shù)列{an}的前n項(xiàng)和為Sn,且(Sn-1)2=anSn(n∈N*),設(shè)bn=(-1)n+1(n+1)2·anan+1(n∈N*),數(shù)列{bn(1)求S1,S2,S3的值;(2)猜想數(shù)列{an}的前n項(xiàng)和Sn,并用數(shù)學(xué)歸納法加以證明;(3)求數(shù)列{Tn}的通項(xiàng)公式.解:(1)由(Sn-1)令n=1,則(S解得S1=12當(dāng)n≥2時(shí),由an=Sn-Sn-1,得(Sn-1)2=(Sn-Sn-1)Sn,得Sn(2-S令n=2,得S2=23令n=3,得S3=34即S1=12,S2=23,S3=(2)由(1)知S1=12,S2=23,S3=34,猜想Sn=nn+1(n下面用數(shù)學(xué)歸納法證明:①當(dāng)n=1時(shí),S1=12,1②假設(shè)當(dāng)n=k(k∈N*)時(shí),猜想成立,即Sk=kk那么當(dāng)n=k+1時(shí),由(1)知Sk+1=12即當(dāng)n=k+1時(shí),猜想也成立.由①②可知,猜想對(duì)于任何n∈N*都成立.(3)由(2)知a1=12.當(dāng)n≥2時(shí),an=Sn-Sn-1=n且a1=12符合上式,即