專(zhuān)題二-數(shù)列-極限-數(shù)學(xué)歸納法

專(zhuān)題二-數(shù)列-極限-__9nn問(wèn)題1數(shù)列{a}滿(mǎn)足a,aaan2a,(nN).n1212nn(I)則{a}的通項(xiàng)公式a=;nnn(III)設(shè)函數(shù)f(n)是1an100n與n的最大者,則f(n)的最小值為.問(wèn)題2已知定義在R上的函數(shù)f(x)和數(shù)列{a}滿(mǎn)足下列條件:nann121nn1nn1baanNbnn1nn(II)求數(shù)列{a}的通項(xiàng)公式;(III)當(dāng)k1時(shí),求lima.nnP于零的等差數(shù)列.00nnnn1kDknnnnT3n1bnnDnDA,(0,2B2C,222n125nn2nn4834343A,tB,tC,tD,t75752575507550112233a2b2A,2x=x+x13B,2y=y+y2131xxx13213為A角三角形B,銳角三角形C,等腰直角三角形D,以上都不對(duì)nnn=.n662636nnn9在等比數(shù)列{a}中,lim(a+a+...+a)=,則a的取值范圍是.nn12n151nnn前2m項(xiàng)之和S=.2m11等差數(shù)列{a}中,S是它的前n項(xiàng)和且SS,S>S,則①此數(shù)列的公差d0,nn67789677n是.3n123nnn又f(1)=n2,f(一1)=n,(I)求數(shù)列的通項(xiàng)a;(II)試比較f()與3的大小,并說(shuō)明理n2.13已知函數(shù)f(x)=3x2+bx+1是偶函數(shù),g(x)=5x+c是奇函數(shù),正數(shù)數(shù)列{a}滿(mǎn)n足1n+1nn+1nn(I)若{a}前n項(xiàng)的和為S,則limS=;nnnnnn+1nxx122<x一x,且存在x,使得f(x)=x,數(shù)列{a}中,a<x,f(a)=2a一a(nN),12000n10nn+1n求證:對(duì)于任意的自然數(shù)n,有:(I)a<x;(II)a<x.n0nn+1參考答案:12nnnnn1aaaaan1于是n2aa45naaaaan1于是n2aa45n1n(n1).12nn(n1).aaa323n11an所以當(dāng)n49或50時(shí),1100n有最小值2450.an(III)因f(n)是100n與n的最大者,有f(n)1100n(100n),nan有f(n)=f(1)=1.min問(wèn)題2(I)證明:由baa0,得baaf(a)f(a)k(aa)0.1212322121由數(shù)學(xué)歸納法可證baa0(nN).nn1nn1nn1nn1nn1因此,數(shù)列是一個(gè)公比為k的等比數(shù)列.nn12112n21而bbb(aa)(aa)(aa)aa(n2),有12n2132nn1n1n121x2y211[2(1x)2(1x)]x2y23(II)設(shè)P(x,y),則由點(diǎn)P在半圓C上知,PMPNx2y2100000000又PMPN0=(1x)2y2(1又PMPN0=0000PM得cos4PM得cos4x22,有cos1,02PN4x200PM1134x2000nn2an2ana1a2n1(2n1)S2n12(2n1)2n1,選B.bn2bnb1b2n1b1b2n1(2n1)T2n13(2n1)13n122515115②當(dāng)0q1時(shí),由aqaq2a,得q1,于是得q,選D.2210191nn2nn2173又a,于是t,選D.12575505由橢圓第2定義得AFCF(xa2)(xa2)2BF2(xa2),選A.1c3c2c3123456nn1n2n3n4n5aa1n2n16n51n1n211n323n2222nnn1111232nn+2nn+2n=10,且a=6.所以{a}中的奇數(shù)項(xiàng)構(gòu)成以a=6為首項(xiàng),公差d=10的等差數(shù)列.n1n+2n項(xiàng)為a與a,由a=2,得a=2,且a=2n22nn+2nan22所以{a}中的偶數(shù)項(xiàng)構(gòu)成以a=2為首項(xiàng),公比q=2的等比數(shù)列.n2m212.m212.677878967ndn12n1n1①12n1nn222222222222222222n+1nn+1nnSn+1nn+1nn+1na3n)wnnnnn+1n1854n3n13424310k010k0k000k0kk00k0