高考數(shù)學(xué)第二輪復(fù)習(xí)數(shù)列典型例題

./1已知數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0滿足：SKIPIF1<0〔a為常數(shù),且SKIPIF1<0．〔Ⅰ求SKIPIF1<0的通項(xiàng)公式；〔Ⅱ設(shè)SKIPIF1<0,若數(shù)列SKIPIF1<0為等比數(shù)列,求a的值；〔Ⅲ在滿足條件〔Ⅱ的情形下,設(shè)SKIPIF1<0,數(shù)列SKIPIF1<0的前n項(xiàng)和為Tn求證：SKIPIF1<0．解：〔ⅠSKIPIF1<0∴SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0SKIPIF1<0,即SKIPIF1<0是等比數(shù)列．∴SKIPIF1<0；〔Ⅱ由〔Ⅰ知,SKIPIF1<0,若SKIPIF1<0為等比數(shù)列,則有SKIPIF1<0而SKIPIF1<0故SKIPIF1<0,解得SKIPIF1<0,再將SKIPIF1<0代入得SKIPIF1<0成立,所以SKIPIF1<0．〔III證明：由〔Ⅱ知SKIPIF1<0,所以SKIPIF1<0SKIPIF1<0SKIPIF1<0,由SKIPIF1<0得SKIPIF1<0所以SKIPIF1<0,從而SKIPIF1<0SKIPIF1<0SKIPIF1<0．即SKIPIF1<0．2數(shù)列SKIPIF1<0中,SKIPIF1<0,SKIPIF1<0〔SKIPIF1<0是常數(shù),SKIPIF1<0,且SKIPIF1<0成公比不為SKIPIF1<0的等比數(shù)列?！睮求SKIPIF1<0的值；〔II求SKIPIF1<0的通項(xiàng)公式。解：〔=1\*ROMANISKIPIF1<0,SKIPIF1<0,SKIPIF1<0,因?yàn)镾KIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,所以SKIPIF1<0,解得SKIPIF1<0或SKIPIF1<0．當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,不符合題意舍去,故SKIPIF1<0．〔=2\*ROMANII當(dāng)SKIPIF1<0時(shí),由于SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,所以SKIPIF1<0。又SKIPIF1<0,SKIPIF1<0,故SKIPIF1<0．當(dāng)n=1時(shí),上式也成立,所以SKIPIF1<03已知數(shù)列SKIPIF1<0中,SKIPIF1<0〔1求證：數(shù)列SKIPIF1<0與SKIPIF1<0都是等比數(shù)列；〔2求數(shù)列SKIPIF1<0前SKIPIF1<0的和SKIPIF1<0；〔3若數(shù)列SKIPIF1<0前SKIPIF1<0的和為SKIPIF1<0,不等式SKIPIF1<0對(duì)SKIPIF1<0恒成立,求SKIPIF1<0的最大值。解：〔1∵SKIPIF1<0,∴SKIPIF1<0∴數(shù)列SKIPIF1<0是以1為首項(xiàng),SKIPIF1<0為公比的等比數(shù)列；數(shù)列SKIPIF1<0是以SKIPIF1<0為首項(xiàng),SKIPIF1<0為公比的等比數(shù)列。 〔2SKIPIF1<0SKIPIF1<0〔3SKIPIF1<0SKIPIF1<0當(dāng)且僅當(dāng)SKIPIF1<0時(shí)取等號(hào),所以SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0的最大值為－484已知等差數(shù)列SKIPIF1<0的公差大于0,且SKIPIF1<0是方程SKIPIF1<0的兩根,數(shù)列SKIPIF1<0的前n項(xiàng)的和為SKIPIF1<0,且SKIPIF1<0.〔1求數(shù)列SKIPIF1<0,SKIPIF1<0的通項(xiàng)公式；〔2記SKIPIF1<0,求證：SKIPIF1<0.解：〔Ⅰ∵a3,a5是方程SKIPIF1<0的兩根,且數(shù)列SKIPIF1<0的公差d>0,∴a3=5,a5=9,公差SKIPIF1<0∴SKIPIF1<0又當(dāng)n=1時(shí),有b1=S1=1－SKIPIF1<0當(dāng)SKIPIF1<0∴數(shù)列{bn}是等比數(shù)列,SKIPIF1<0∴SKIPIF1<0〔Ⅱ由〔Ⅰ知SKIPIF1<0∴SKIPIF1<0∴SKIPIF1<05已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,對(duì)一切正整數(shù)SKIPIF1<0,點(diǎn)SKIPIF1<0都在函數(shù)SKIPIF1<0的圖像上,且過(guò)點(diǎn)SKIPIF1<0的切線的斜率為SKIPIF1<0．〔1求數(shù)列SKIPIF1<0的通項(xiàng)公式．〔2若SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0．〔3設(shè)SKIPIF1<0,等差數(shù)列SKIPIF1<0的任一項(xiàng)SKIPIF1<0,其中SKIPIF1<0是SKIPIF1<0中的最小數(shù),SKIPIF1<0,求SKIPIF1<0的通項(xiàng)公式.解：〔1SKIPIF1<0點(diǎn)SKIPIF1<0都在函數(shù)SKIPIF1<0的圖像上,SKIPIF1<0SKIPIF1<0,當(dāng)SKIPIF1<0時(shí),SKIPIF1<0當(dāng)ｎ＝1時(shí),SKIPIF1<0滿足上式,所以數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0〔2由SKIPIF1<0求導(dǎo)可得SKIPIF1<0SKIPIF1<0過(guò)點(diǎn)SKIPIF1<0的切線的斜率為SKIPIF1<0,SKIPIF1<0.SKIPIF1<0.SKIPIF1<0①由①×4,得SKIPIF1<0②①-②得：SKIPIF1<0SKIPIF1<0SKIPIF1<0〔3SKIPIF1<0,SKIPIF1<0.又SKIPIF1<0,其中SKIPIF1<0是SKIPIF1<0中的最小數(shù),SKIPIF1<0.SKIPIF1<0是公差是4的倍數(shù),SKIPIF1<0.又SKIPIF1<0,SKIPIF1<0,解得ｍ＝27.所以SKIPIF1<0,設(shè)等差數(shù)列的公差為SKIPIF1<0,則SKIPIF1<0SKIPIF1<0,所以SKIPIF1<0的通項(xiàng)公式為SKIPIF1<06已知SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和,SKIPIF1<0,且SKIPIF1<0,其中SKIPIF1<0.<1>求數(shù)列SKIPIF1<0的通項(xiàng)公式SKIPIF1<0；<2>求SKIPIF1<0.解：①SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0又SKIPIF1<0也滿足上式,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0〔SKIPIF1<0SKIPIF1<0數(shù)列SKIPIF1<0是公比為2,首項(xiàng)為SKIPIF1<0的等比數(shù)列SKIPIF1<0②SKIPIF1<0SKIPIF1<0②SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<07函數(shù)SKIPIF1<0對(duì)任意x∈R都有f<x>＋f<1－x>＝EQ\f<1,2>.〔1求SKIPIF1<0的值；〔2數(shù)列SKIPIF1<0的通項(xiàng)公式?！?令SKIPIF1<0試比較Tn與Sn的大小。解：〔1令SKIPIF1<0令SKIPIF1<0〔2SKIPIF1<0又SKIPIF1<0,兩式相加SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0是等差數(shù)列〔3SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<08、已知數(shù)列SKIPIF1<0中SKIPIF1<0,其前n項(xiàng)和為滿足SKIPIF1<0．〔1試求數(shù)列SKIPIF1<0的通項(xiàng)公式．〔2令SKIPIF1<0SKIPIF1<0是數(shù)列SKIPIF1<0的前n項(xiàng)和,證明：SKIPIF1<0．〔3證明:對(duì)任意的SKIPIF1<0,均存在SKIPIF1<0,使得〔2中的SKIPIF1<0成立．解：〔1由SKIPIF1<0得SKIPIF1<0SKIPIF1<0,SKIPIF1<0,即SKIPIF1<0又SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0故數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0．〔2SKIPIF1<0SKIPIF1<0SKIPIF1<0〔3證明：由〔2可知SKIPIF1<0若SKIPIF1<0,則得SKIPIF1<0,化簡(jiǎn)得SKIPIF1<0SKIPIF1<0,SKIPIF1<0當(dāng)SKIPIF1<0,即SKIPIF1<0當(dāng)SKIPIF1<0,即SKIPIF1<0SKIPIF1<0,取SKIPIF1<0即可,綜上可知,對(duì)任意的SKIPIF1<0均存在SKIPIF1<0使得時(shí)〔2中的SKIPIF1<0成立9已知數(shù)列{an}的前n項(xiàng)和為Sn,并且滿足a1＝2,nan＋1＝Sn＋n<n＋1>.〔1求數(shù)列SKIPIF1<0；〔2設(shè)SKIPIF1<0解：〔1SKIPIF1<0SKIPIF1<0〔2SKIPIF1<0SKIPIF1<010已知二次函數(shù)f<x>＝ax2＋bx＋c的圖象頂點(diǎn)坐標(biāo)是<EQ\f<3,2>,－EQ\f<1,4>>,且f<3>＝2〔1求y＝f<x>的表達(dá)式,并求出f<1>,f<2>的值；〔2數(shù)列SKIPIF1<0,若對(duì)任意的實(shí)數(shù)SKIPIF1<0,其中SKIPIF1<0是定義在實(shí)數(shù)集R上的一個(gè)函數(shù),求數(shù)列SKIPIF1<0的通項(xiàng)公式；解：〔1SKIPIF1<0SKIPIF1<0〔2令SKIPIF1<0SKIPIF1<011已知數(shù)列{an}滿足a1＝5,a2＝5,an＋1＝an＋6an－1<n≥2且n∈N*>〔1求出所有使數(shù)列SKIPIF1<0值,并說(shuō)明理由；〔2求數(shù)列SKIPIF1<0的通項(xiàng)公式；〔3求證：SKIPIF1<0解：〔1SKIPIF1<0〔2SKIPIF1<0〔3當(dāng)SKIPIF1<0時(shí),SKIPIF1<0SKIPIF1<0.〔Ⅰ求數(shù)列SKIPIF1<0的通項(xiàng)公式；〔Ⅱ求證：SKIPIF1<0；〔Ⅲ求證：當(dāng)SKIPIF1<0時(shí),SKIPIF1<0．解：〔1由SKIPIF1<0,得SKIPIF1<0,代入SKIPIF1<0,得SKIPIF1<0,整理,得SKIPIF1<0,從而有SKIPIF1<0,SKIPIF1<0,SKIPIF1<0是首項(xiàng)為1,公差為1的等差數(shù)列,SKIPIF1<0即SKIPIF1<0.〔2SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0.〔3SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0.由〔2知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0.13已知數(shù)列SKIPIF1<0的首項(xiàng)SKIPIF1<0,前SKIPIF1<0項(xiàng)和為SKIPIF1<0,且SKIPIF1<0、SKIPIF1<0、SKIPIF1<0分別是直線SKIPIF1<0上的點(diǎn)A、B、C的橫坐標(biāo),點(diǎn)B分SKIPIF1<0所成的比為SKIPIF1<0,設(shè)SKIPIF1<0SKIPIF1<0。⑴判斷數(shù)列SKIPIF1<0是否為等比數(shù)列,并證明你的結(jié)論；⑵設(shè)SKIPIF1<0,證明：SKIPIF1<0。⑴由題意得SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0數(shù)列SKIPIF1<0是以SKIPIF1<0為首項(xiàng),以2為公比的等比數(shù)列。[則SKIPIF1<0SKIPIF1<0SKIPIF1<0〔SKIPIF1<0]⑵由SKIPIF1<0及SKIPIF1<0得SKIPIF1<0SKIPIF1<0SKIPIF1<0則SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<014已知各項(xiàng)均為正數(shù)的數(shù)列SKIPIF1<0滿足SKIPIF1<0且SKIPIF1<0是SKIPIF1<0、SKIPIF1<0的等差中項(xiàng)〔1求數(shù)列SKIPIF1<0的通項(xiàng)公式SKIPIF1<0；〔2若SKIPIF1<0,求使SKIPIF1<0成立的正整數(shù)SKIPIF1<0的最小值。解：SKIPIF1<0SKIPIF1<0數(shù)列SKIPIF1<0的各項(xiàng)均為正數(shù),SKIPIF1<0SKIPIF1<0,即SKIPIF1<0SKIPIF1<0數(shù)列SKIPIF1<0是以2為公比的等比數(shù)列。SKIPIF1<0SKIPIF1<0是SKIPIF1<0的等差中項(xiàng),SKIPIF1<0SKIPIF1<0SKIPIF1<0數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0〔2由〔1及SKIPIF1<0,得SKIPIF1<0,〔6分SKIPIF1<0①SKIPIF1<0②②-①得,SKIPIF1<0要使SKIPIF1<0成立,只需SKIPIF1<0成立,即SKIPIF1<0SKIPIF1<0成立的正整數(shù)n的最小值為5。15已知SKIPIF1<0,且SKIPIF1<0,數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0,它滿足條件SKIPIF1<0.數(shù)列SKIPIF1<0中,SKIPIF1<0。〔1求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0；〔2若對(duì)一切SKIPIF1<0都有SKIPIF1<0,求SKIPIF1<0的取值范圍。解：〔1SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0.當(dāng)SKIPIF1<0≥2時(shí),SKIPIF1<0=SKIPIF1<0,SKIPIF1<0SKIPIF1<0此時(shí)SKIPIF1<0SKIPIF1<0·SKIPIF1<0=SKIPIF1<0·SKIPIF1<0,SKIPIF1<0SKIPIF1<0……SKIPIF1<0=SKIPIF1<0設(shè)SKIPIF1<0……+SKIPIF1<0,SKIPIF1<0SKIPIF1<0……＋SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0SKIPIF1<0〔2由SKIPIF1<0可得SKIPIF1<0當(dāng)SKIPIF1<0時(shí),由SKIPIF1<0可得SKIPIF1<0,SKIPIF1<0SKIPIF1<0SKIPIF1<0對(duì)一切SKIPIF1<0都成立,SKIPIF1<0此時(shí)的解為SKIPIF1<0.SKIPIF1<0當(dāng)SKIPIF1<0時(shí),由SKIPIF1<0可得SKIPIF1<0SKIPIF1<0SKIPIF1<0≥SKIPIF1<0SKIPIF1<0SKIPIF1<0對(duì)一切SKIPIF1<0都成立,SKIPIF1<0此時(shí)的解為SKIPIF1<0.由SKIPIF1<0,SKIPIF1<0可知,對(duì)一切SKIPIF1<0都有SKIPIF1<0的SKIPIF1<0的取值范圍是SKIPIF1<0或SKIPIF1<0.16設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為Sn=2n2,SKIPIF1<0為等比數(shù)列,且SKIPIF1<0〔1求數(shù)列SKIPIF1<0和SKIPIF1<0的通項(xiàng)公式；〔2設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0。解：〔1：當(dāng)SKIPIF1<0SKIPIF1<0故{an}的通項(xiàng)公式為SKIPIF1<0的等差數(shù)列.設(shè){bn}的通項(xiàng)公式為SKIPIF1<0故SKIPIF1<0〔2SKIPIF1<0SKIPIF1<0兩式相減得SKIPIF1<017設(shè)不等式SKIPIF1<0所表示的平面區(qū)域?yàn)镾KIPIF1<0,記SKIPIF1<0內(nèi)的格點(diǎn)〔SKIPIF1<0,SKIPIF1<0〔SKIPIF1<0、SKIPIF1<0∈z的個(gè)數(shù)為SKIPIF1<0〔SKIPIF1<0∈SKIPIF1<0.〔Ⅰ求SKIPIF1<0,SKIPIF1<0的值及SKIPIF1<0的表達(dá)式；〔Ⅱ記SKIPIF1<0,若對(duì)于任意SKIPIF1<0∈SKIPIF1<0,總有SKIPIF1<0≤m成立,求實(shí)數(shù)m的取值范圍；〔Ⅲ設(shè)SKIPIF1<0為數(shù)列｛SKIPIF1<0｝的前SKIPIF1<0項(xiàng)和,其中SKIPIF1<0＝SKIPIF1<0,問(wèn)是否存在正整數(shù)SKIPIF1<0、t,使SKIPIF1<0＜SKIPIF1<0成立？若存在,求出正整數(shù)SKIPIF1<0,t；若不存在,請(qǐng)說(shuō)明理由.解：〔ⅠSKIPIF1<0＝3,SKIPIF1<0＝6.由SKIPIF1<0＞0,0＜SKIPIF1<0≤SKIPIF1<0,得0＜SKIPIF1<0＜3,又SKIPIF1<0∈SKIPIF1<0,∴SKIPIF1<0＝1,或SKIPIF1<0＝2.當(dāng)SKIPIF1<0＝1,0＜SKIPIF1<0≤2SKIPIF1<0時(shí),共有2SKIPIF1<0個(gè)格點(diǎn)；當(dāng)SKIPIF1<0＝2,0＜SKIPIF1<0≤SKIPIF1<0時(shí),共有SKIPIF1<0個(gè)格點(diǎn).故SKIPIF1<0.〔Ⅱ由〔1知SKIPIF1<0＝SKIPIF1<0,則SKIPIF1<0－SKIPIF1<0＝SKIPIF1<0.∴當(dāng)SKIPIF1<0≥3時(shí),SKIPIF1<0＜SKIPIF1<0.又SKIPIF1<0＝9＜SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0,所以SKIPIF1<0≤SKIPIF1<0,故SKIPIF1<0≥SKIPIF1<0.〔Ⅲ假設(shè)存在滿足題意的SKIPIF1<0和SKIPIF1<0,由〔1知SKIPIF1<0＝SKIPIF1<0＝SKIPIF1<0,故SKIPIF1<0.則SKIPIF1<0＜SKIPIF1<0.變形得SKIPIF1<0＜SKIPIF1<0,即SKIPIF1<0＜0.∴1＜SKIPIF1<0〔8－SKIPIF1<0＜15.由于SKIPIF1<0、SKIPIF1<0均為正整數(shù),所以SKIPIF1<0＝SKIPIF1<0＝1.附:SKIPIF1<0,SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),由SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0.當(dāng)SKIPIF1<0時(shí),SKIPIF1<0,由SKIPIF1<0,得SKIPIF1<0,SKIPIF1<0不存在.所以SKIPIF1<0＝SKIPIF1<0＝1.18已知二次函數(shù)SKIPIF1<0的圖像經(jīng)過(guò)坐標(biāo)原點(diǎn),其導(dǎo)函數(shù)為SKIPIF1<0數(shù)列{SKIPIF1<0}的前n項(xiàng)和為SKIPIF1<0,點(diǎn)SKIPIF1<0均在函數(shù)SKIPIF1<0的圖像上.〔I求數(shù)列{SKIPIF1<0}的通項(xiàng)公式；〔II設(shè)SKIPIF1<0,SKIPIF1<0的前n項(xiàng)和,求使得SKIPIF1<0對(duì)所有SKIPIF1<0都成立的最小正整數(shù)m.解：〔I設(shè)這二次函數(shù)SKIPIF1<0,由于SKIPIF1<0,得SKIPIF1<0又因?yàn)辄c(diǎn)SKIPIF1<0的圖像上,所以SKIPIF1<0當(dāng)SKIPIF1<0SKIPIF1<0〔II由〔I得知SKIPIF1<0SKIPIF1<0故SKIPIF1<0SKIPIF1<0因此,要使SKIPIF1<0,必須且僅須滿足SKIPIF1<0即SKIPIF1<0,所以滿足要求的最小正整數(shù)m為10。19數(shù)列SKIPIF1<0,由下列條件確定：①a1＜0,b1＜0.②當(dāng)k≥2時(shí),ak和bk滿足下列條件：當(dāng)SKIPIF1<0. 〔1若SKIPIF1<0,SKIPIF1<0,分別寫出{an}、{bn}的前四項(xiàng). 〔2證明數(shù)列{ak－bk}是等比數(shù)列. 〔3設(shè)SKIPIF1<0是滿足b1＞b2＞…＞bn的最大整數(shù)時(shí),用a1、b1表示n滿足的條件.解：〔1SKIPIF1<0SKIPIF1<0〔2當(dāng)SKIPIF1<0時(shí),SKIPIF1<0當(dāng)SKIPIF1<0時(shí),SKIPIF1<0又SKIPIF1<0,∴數(shù)列SKIP