數(shù)列放縮通項(xiàng)證明不等式與數(shù)列不等式恒成立問題數(shù)列通項(xiàng)放縮問題是放縮問題的常考類型，相較于求和之后再些常見的裂項(xiàng)公式與放縮公式需要注意. 3題型二與導(dǎo)數(shù)結(jié)合的放縮 8 91<<n2(n n?2bn?3b211n+?n+?n（1）<=?n≥2)；（3）=<=2?；（4）Tr+1=C（5）=<=2?+?+；（8）=<= n+1?n?1 1 n+1+n?122<nn?1+n+n?1)=?22n?12n?1<2n?12n?2=2n?1n?1?1=?n≥2)；?1n<1+1+++…+<3；?12n?12n?1=?n≥2). ++（14）2(?)=<< ++nn+1=?.記數(shù)列{an}的前n項(xiàng)和為Sn，則()2<S<3B．3<S100<4C．4<S<922<S<5n+1= n 1+a21?an+1=an+an=|2an+1（an)an+1an∴<|+2an+1（an)an+1an,即1<1<2aan+1根據(jù)累加法可得，1n?1n+1≤1+=nn4(n+1)2n+1∴an+1 n≤ ≤1+aan+1 n=a2n+32n+3n.n+1 >?(n+1)2n+1n+2n+1n+3a n+1n+1n+3a n+1≤6(n+1)(n+2),S≤6?+?+?+…+?=6?<3，即<S100<3．故選：A.題型=通項(xiàng)放縮，2=42<42?1==2．已知an=n+1+32 ++++…<【詳解】a33n+133n3n+1 a33n+133n3n+1n所以1+1+1+…+1<2+2+2+…+2=1?=1?1<1a1a2a3an3233343n+1a1a2an2a1a2an2n?.1n?1an?bn=(a?b)(an?1+an?2b+an?3b2+…+abn?2+bn?1).nT<n+1.n【詳解】Sn=n2，則Sn+1=(n+1)2，b=1+.b=2n2+2b=2,則b=nnn2+2n+3.n+1+nn+nnb?1=n2+2n+3?1=n2+2n+3?(n+1)nn+1n+12=<=<2= <?2nn+1.nnn+1∴T=b1+b2+…+b<n+1?1<nnn+1 ++++ ++++【答案】B<=+?=，11>= n+n+2 ?=n+n+2n+2?n)= ?據(jù)此可得答案.【詳解】+?=， 1+23+45+6 1+23+45+699+100又n+n+1>n+n+2=(n+n+2n+2?n=2,則1+2+3+4+5+6+…+99+100>2=?+?+?+…+?)= 2 >=.21+23+45+699+10021+23+45+699+100kk+12k2kcc4k2?1kk+12k2kcc4k2?1kkk+12k2kccS=2?1?n<22n?12n<S<2,故得6．已知正項(xiàng)數(shù)列{an}的前n項(xiàng)和為Sn，且滿足a=2anSn?1.lSnJ?n?Sn?12=2所以數(shù)列（2）S=2S1S1?1?S=1，由（1）可知數(shù)列{ =2(n+1?n)，=2(n+1?n)，==>Sn2nn+n+1 n T>2(?1)+2(?)+…+2(101?100)=2(?1)>2(?1)=b?b=nb?b=c=bc=b,證明：n2n【答案】(1)【答案】(1)（2）由放縮法與錯(cuò)位相減法求和證明.S8=8a1+dS8=8a1+d=64b3?b2=48=4(q2?q),而,解得,故n}nnq=4d=2,則,而公比,解得,故,則,而公比,解得,故=4q>0(2k?1)(2k+1)<1+4(2k?1)(2k+1)<1+4n==,則142k4k144k44k+2k?(4+2k)+4S=S=++…+,則令 222232n+1 22222n2n+12n2n+1kkk+12k2kccnn+1nn+1nn23n+1n+1nn23n+1n+1所以S=T1+T2+…+T=又因?yàn)閍n+1?=?=?,2n+222n+2所以Sn>an+1?.9．已知an4n=n,數(shù)列bn=nan?1，證明： b1 ≤bn1n?11 ≤bn1n?1公式可證得原不等式成立，綜合可得出結(jié)論.,再結(jié)合等比數(shù)列求和nn?=3<9，b,492n+122n+12【解析】n=2n+12n+1?32n+12n+1?332n+1?3,32n+1?332n+1?3??3?3?24?3+24?33?3?24?3+24?3?25?3+…+2n?3?2n+1?3)|=12+24253?3?3?2n+1?3)246 + 255=12+5425nn題型二與導(dǎo)數(shù)結(jié)合的放縮1n+11n+1112017全國3卷）已知函數(shù)f(x)=x?1?alnx.2n2n2nln(1+2)+ln(1+22)+n)<2+22+n=n)<e,而(1+2)(1+22)(1+23)>2，所以m的最小值為3. b?a>2 +當(dāng)b>b?a>2 +,即lnb?lna<(+)b?a.2ab令a=n,b=n+1，則ln(n+1)?lnn<(+)，所以ln(n+1)?lnn<(+)①.2nn+12nn+1若再利用<L(a,b) abbbaxb?lna?lnb<?ln<??2lnx<x abbbaxb+n 1+n 1令t=1+n1111 n1+n12n2n+n故 2nn+4n+4n2+4n2212n2n+n1+...+...+22++ 22++12n2n+n題型三數(shù)列恒成立問題【分析】設(shè)等差數(shù)列{an}的公差為d，由已知可得d=?2，求得Sn，由數(shù)列的單調(diào)性列不等式即可得a1的【詳解】設(shè)等差數(shù)列{an}的公差為d，由于3a2+2a3=S5+6，所以3(a1+d)+2(a1+2d)=5a1+10d+6，解得d=?2，所以Sn=na1+d=?n2+(a1+1)n，n12n?1=【分析】由a1=1，2an+1=a12n?1=nnn2?3n?22n?1,結(jié)合bn+1?bn=?，分1≤n≤52?3n+1n2n2n?12nn+1?bn≥0，∴bn+1≥bn，當(dāng)且僅當(dāng)n=5時(shí)取等號(hào)，b6=b5，max==， 1,2【答案】7所以λan≤a+12?λn≤n2+12?λ≤n+，min=7所以λ≤7)【分析】先由題設(shè)求得an，然后利用數(shù)列的單調(diào)性求得其最大值，把對任意λ>0，所有的正整數(shù)n都有到答案.22a322a3nnnn+，即k<λ+對任意λ>0恒成立，由λ+≥2=，當(dāng)且僅當(dāng)λ=,即λ=時(shí)取最小值，則k.3,218．已知an=3n2+n，若an≤λ2n對于任意n∈N*恒成立，則實(shí)數(shù)λ的取值范圍是.【分析】先分離參數(shù)將問題轉(zhuǎn)化為≤λ對于任意n∈N*恒成立，進(jìn)而轉(zhuǎn)化為()max≤λ,構(gòu)造b=n3n2+n2n,再作差判定單調(diào)性求出數(shù)列{bn}的最值，進(jìn)而求出λ的取值范圍.【詳解】因?yàn)閍n=3n2+n，且an≤λ2n對于任意n∈N*恒成立，2n≤λ對于任意n∈N*恒成立，即()max≤λ,令bn=，則bn+1?bn=?=，因?yàn)閎2?b1=>0，b3?b2=>0，b4?b3且bn+1?bn=<0對于任意n≥3=?<0,2+n)max=b3=15，2n4所以實(shí)數(shù)λ的取值范圍是3 a2n3 a2n值為()+12n2+n ?+12n2+n 13 16 19 【答案】Dn?3an?1=4n，故代入an≥整理得2≥，利用作差法得到單調(diào)遞減，??3n+1n=Sn?Sn?12nn?1?3a+n?12,方程兩邊同除以3n，得?=4，n≥，n+1nn+1?3n1?2nn+1n1?2nn+13n3n+13n+13n+3n3n+ n+13n3n+13n+13n+3n3n+n+1nn+3nn+3n則k的最小值為解決不等式恒成立問題即可.n}是等差數(shù)列，設(shè)其公差為d，解方程x2?18x+65=0得x=5或x=13，又a7>a3，則bn+1?bn== n22n?1?n?2=?2n3+5n2