新高考數(shù)學(xué)二輪復(fù)習(xí)對(duì)點(diǎn)題型第17講數(shù)列的通項(xiàng)、求和及數(shù)列不等式的證明（學(xué)生版）

第17講數(shù)列的通項(xiàng)、求和及數(shù)列不等式的證明真題展示2022新高考一卷第17題記SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，已知SKIPIF1<0，SKIPIF1<0是公差為SKIPIF1<0的等差數(shù)列．（1）求SKIPIF1<0的通項(xiàng)公式；（2）證明：SKIPIF1<0．試題亮點(diǎn)試題以考生熟悉的等差數(shù)列為載體而設(shè)計(jì)，但不是通常的給定等差數(shù)列求通項(xiàng)、求和等常規(guī)操作，而是將等差數(shù)列的性質(zhì)融合在前n項(xiàng)和與通項(xiàng)的關(guān)系之中，特別是第（2）問中的數(shù)列的求和運(yùn)算涉及裂項(xiàng)相消．試題源于教材、其創(chuàng)新思想又高于教材，充分體現(xiàn)高考的選拔功能.試題對(duì)高中數(shù)學(xué)教學(xué)具有指導(dǎo)作用，要求考生在強(qiáng)化基本功的同時(shí)，加強(qiáng)對(duì)知識(shí)的靈活運(yùn)用，形成學(xué)科素養(yǎng).知識(shí)要點(diǎn)整理數(shù)列求和問題數(shù)列求和是數(shù)列問題中的基本題型，是數(shù)列部分的重點(diǎn)內(nèi)容，在高考中也占據(jù)重要地位，它具有復(fù)雜多變、綜合性強(qiáng)、解法靈活等特點(diǎn)．?dāng)?shù)列求和的方法主要有公式法、分組轉(zhuǎn)化法、倒序相加法、錯(cuò)位相減法、裂項(xiàng)相消法、并項(xiàng)求和法等．一、公式法求和例1求數(shù)列1,3＋5,7＋9＋11,13＋15＋17＋19，…的前n項(xiàng)和．反思感悟公式法求和中的常用公式有(1)等差、等比數(shù)列的前n項(xiàng)和①等差數(shù)列：Sn＝na1＋eq\f(nn－1,2)d(d為公差)或Sn＝eq\f(na1＋an,2).②等比數(shù)列：Sn＝eq\b\lc\{\rc\(\a\vs4\al\co1(na1，q＝1，,\f(a11－qn,1－q)＝\f(a1－anq,1－q)，q≠1，))其中q為公比．(2)四類特殊數(shù)列的前n項(xiàng)和①1＋2＋3＋…＋n＝eq\f(1,2)n(n＋1)．②1＋3＋5＋…＋(2n－1)＝n2.③12＋22＋32＋…＋n2＝eq\f(1,6)n(n＋1)(2n＋1)．④13＋23＋33＋…＋n3＝eq\f(1,4)n2(n＋1)2.二、分組轉(zhuǎn)化法求和例2求和：Sn＝eq\b\lc\(\rc\)(\a\vs4\al\co1(x＋\f(1,x)))2＋eq\b\lc\(\rc\)(\a\vs4\al\co1(x2＋\f(1,x2)))2＋…＋eq\b\lc\(\rc\)(\a\vs4\al\co1(xn＋\f(1,xn)))2(x≠0)．反思感悟某些數(shù)列，通過適當(dāng)分組，可得出兩個(gè)或幾個(gè)等差數(shù)列或等比數(shù)列，進(jìn)而利用等差數(shù)列或等比數(shù)列的求和公式分別求和，從而得出原數(shù)列的和．三、倒序相加法求和例3設(shè)F(x)＝eq\f(4x,4x＋2)，求Feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2021)))＋Feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2,2021)))＋…＋Feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2020,2021))).反思感悟(1)倒序相加法類比推導(dǎo)等差數(shù)列的前n項(xiàng)和公式時(shí)所用的方法，就是將一個(gè)數(shù)列倒過來排列(反序)，再把它與原數(shù)列相加，就可以得到n個(gè)(a1＋an)．(2)如果一個(gè)數(shù)列{an}，首末兩端等“距離”的兩項(xiàng)的和相等，那么求其和可以用倒序相加法．四、裂項(xiàng)相消法求和例4求和：eq\f(1,22－1)＋eq\f(1,32－1)＋eq\f(1,42－1)＋…＋eq\f(1,n2－1)，n≥2，n∈N*.延伸探究求和：eq\f(22,22－1)＋eq\f(32,32－1)＋eq\f(42,42－1)＋…＋eq\f(n2,n2－1)，n≥2，n∈N*.反思感悟(1)對(duì)于裂項(xiàng)后明顯有能夠相消的項(xiàng)的一類數(shù)列，在求和時(shí)常用“裂項(xiàng)法”，分式的求和多利用此法，可用待定系數(shù)法對(duì)通項(xiàng)公式拆項(xiàng)，相消時(shí)應(yīng)注意消去項(xiàng)的規(guī)律，即消去哪些項(xiàng)，保留哪些項(xiàng)．(2)常見的拆項(xiàng)公式有①eq\f(1,nn＋1)＝eq\f(1,n)－eq\f(1,n＋1).②eq\f(1,nn＋k)＝eq\f(1,k)eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,n)－\f(1,n＋k))).③eq\f(1,2n－12n＋1)＝eq\f(1,2)eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2n－1)－\f(1,2n＋1))).④eq\f(1,\r(n)＋\r(n＋1))＝eq\r(n＋1)－eq\r(n).⑤eq\f(1,nn＋1n＋2)＝eq\f(1,2)eq\b\lc\[\rc\](\a\vs4\al\co1(\f(1,nn＋1)－\f(1,n＋1n＋2))).五、錯(cuò)位相減法求和例5已知{an}是等比數(shù)列，{bn}是等差數(shù)列，且a1＝1，b1＝3，a2＋b2＝7，a3＋b3＝11.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式；(2)設(shè)cn＝eq\f(bn,an)，n∈N*，求數(shù)列{cn}的前n項(xiàng)和Tn.反思感悟一般地，如果數(shù)列{an}是等差數(shù)列，{bn}是等比數(shù)列，求數(shù)列{an·bn}的前n項(xiàng)和時(shí)，可采用錯(cuò)位相減法求和，在寫出“Sn”與“qSn”的表達(dá)式時(shí)應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對(duì)齊”以便于下一步準(zhǔn)確寫出“Sn－qSn”的表達(dá)式．六、并項(xiàng)求和法求和例6求和：Sn＝－1＋3－5＋7－…＋(－1)n(2n－1)．反思感悟通項(xiàng)中含有(－1)n的數(shù)列求前n項(xiàng)和時(shí)可以考慮使用奇偶并項(xiàng)法，分項(xiàng)數(shù)為奇數(shù)和偶數(shù)分別進(jìn)行求和．三年真題1．設(shè)SKIPIF1<0是等差數(shù)列，SKIPIF1<0是等比數(shù)列，且SKIPIF1<0．(1)求SKIPIF1<0與SKIPIF1<0的通項(xiàng)公式；(2)設(shè)SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，求證：SKIPIF1<0；(3)求SKIPIF1<0．2．已知SKIPIF1<0為等差數(shù)列，SKIPIF1<0是公比為2的等比數(shù)列，且SKIPIF1<0．(1)證明：SKIPIF1<0；(2)求集合SKIPIF1<0中元素個(gè)數(shù)．3．已知等差數(shù)列SKIPIF1<0的首項(xiàng)SKIPIF1<0，公差SKIPIF1<0．記SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0．(1)若SKIPIF1<0，求SKIPIF1<0；(2)若對(duì)于每個(gè)SKIPIF1<0，存在實(shí)數(shù)SKIPIF1<0，使SKIPIF1<0成等比數(shù)列，求d的取值范圍．4．已知函數(shù)SKIPIF1<0．(1)當(dāng)SKIPIF1<0時(shí)，討論SKIPIF1<0的單調(diào)性；(2)當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0，求a的取值范圍；(3)設(shè)SKIPIF1<0，證明：SKIPIF1<0．5．記SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和．已知SKIPIF1<0．(1)證明：SKIPIF1<0是等差數(shù)列；(2)若SKIPIF1<0成等比數(shù)列，求SKIPIF1<0的最小值．6．已知SKIPIF1<0為有窮整數(shù)數(shù)列．給定正整數(shù)m，若對(duì)任意的SKIPIF1<0，在Q中存在SKIPIF1<0，使得SKIPIF1<0，則稱Q為SKIPIF1<0連續(xù)可表數(shù)列．(1)判斷SKIPIF1<0是否為SKIPIF1<0連續(xù)可表數(shù)列？是否為SKIPIF1<0連續(xù)可表數(shù)列？說明理由；(2)若SKIPIF1<0為SKIPIF1<0連續(xù)可表數(shù)列，求證：k的最小值為4；(3)若SKIPIF1<0為SKIPIF1<0連續(xù)可表數(shù)列，且SKIPIF1<0，求證：SKIPIF1<0．7．記SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和，已知SKIPIF1<0是公差為SKIPIF1<0的等差數(shù)列．(1)求SKIPIF1<0的通項(xiàng)公式；(2)證明：SKIPIF1<0．8．已知SKIPIF1<0是公差為2的等差數(shù)列，其前8項(xiàng)和為64．SKIPIF1<0是公比大于0的等比數(shù)列，SKIPIF1<0．（I）求SKIPIF1<0和SKIPIF1<0的通項(xiàng)公式；（II）記SKIPIF1<0，（i）證明SKIPIF1<0是等比數(shù)列；（ii）證明SKIPIF1<09．記SKIPIF1<0是公差不為0的等差數(shù)列SKIPIF1<0的前n項(xiàng)和，若SKIPIF1<0．（1）求數(shù)列SKIPIF1<0的通項(xiàng)公式SKIPIF1<0；（2）求使SKIPIF1<0成立的n的最小值．10．設(shè)p為實(shí)數(shù).若無窮數(shù)列SKIPIF1<0滿足如下三個(gè)性質(zhì)，則稱SKIPIF1<0為SKIPIF1<0數(shù)列：①SKIPIF1<0，且SKIPIF1<0；②SKIPIF1<0；③SKIPIF1<0，SKIPIF1<0．（1）如果數(shù)列SKIPIF1<0的前4項(xiàng)為2，-2，-2，-1，那么SKIPIF1<0是否可能為SKIPIF1<0數(shù)列？說明理由；（2）若數(shù)列SKIPIF1<0是SKIPIF1<0數(shù)列，求SKIPIF1<0；（3）設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0.是否存在SKIPIF1<0數(shù)列SKIPIF1<0，使得SKIPIF1<0恒成立？如果存在，求出所有的p；如果不存在，說明理由．11．記SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和，已知SKIPIF1<0，且數(shù)列SKIPIF1<0是等差數(shù)列，證明：SKIPIF1<0是等差數(shù)列.12．已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0.（1）求數(shù)列SKIPIF1<0的通項(xiàng)；（2）設(shè)數(shù)列SKIPIF1<0滿足SKIPIF1<0，記SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，若SKIPIF1<0對(duì)任意SKIPIF1<0恒成立，求實(shí)數(shù)SKIPIF1<0的取值范圍.13．記SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和，SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)積，已知SKIPIF1<0．（1）證明：數(shù)列SKIPIF1<0是等差數(shù)列;（2）求SKIPIF1<0的通項(xiàng)公式．14．已知數(shù)列SKIPIF1<0的各項(xiàng)均為正數(shù)，記SKIPIF1<0為SKIPIF1<0的前n項(xiàng)和，從下面①②③中選取兩個(gè)作為條件，證明另外一個(gè)成立．①數(shù)列SKIPIF1<0是等差數(shù)列：②數(shù)列SKIPIF1<0是等差數(shù)列；③SKIPIF1<0．注：若選擇不同的組合分別解答，則按第一個(gè)解答計(jì)分．15．設(shè)SKIPIF1<0是首項(xiàng)為1的等比數(shù)列，數(shù)列SKIPIF1<0滿足SKIPIF1<0．已知SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列．（1）求SKIPIF1<0和SKIPIF1<0的通項(xiàng)公式；（2）記SKIPIF1<0和SKIPIF1<0分別為SKIPIF1<0和SKIPIF1<0的前n項(xiàng)和．證明：SKIPIF1<0．16．已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0（1）記SKIPIF1<0，寫出SKIPIF1<0，SKIPIF1<0，并求數(shù)列SKIPIF1<0的通項(xiàng)公式；（2）求SKIPIF1<0的前20項(xiàng)和.三年模擬一、單選題1．已知角SKIPIF1<0的終邊不在坐標(biāo)軸上，則下列一定成等比數(shù)列的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<02．已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，首項(xiàng)SKIPIF1<0，且滿足SKIPIF1<0，則SKIPIF1<0的值為（

）A．4093 B．4094 C．4095 D．40963．已知數(shù)列SKIPIF1<0為等差數(shù)列，SKIPIF1<0為等比數(shù)列SKIPIF1<0的前n項(xiàng)和，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、填空題4．設(shè)SKIPIF1<0是由正整數(shù)組成且項(xiàng)數(shù)為SKIPIF1<0的增數(shù)列，已知SKIPIF1<0，SKIPIF1<0，數(shù)列SKIPIF1<0任意相鄰兩項(xiàng)的差的絕對(duì)值不超過1，若對(duì)于SKIPIF1<0中任意序數(shù)不同的兩項(xiàng)SKIPIF1<0和SKIPIF1<0，在剩下的項(xiàng)中總存在序數(shù)不同的兩項(xiàng)SKIPIF1<0和SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的最小值為___________.5．已知項(xiàng)數(shù)為m的有限數(shù)列SKIPIF1<0是1，2，3，…，m的一個(gè)排列.若SKIPIF1<0，且SKIPIF1<0，則所有可能的m值之和為______.6．?dāng)?shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0__________7．已知等差數(shù)列SKIPIF1<0中，SKIPIF1<0，則SKIPIF1<0的值等于__________.8．已知公差為SKIPIF1<0且各項(xiàng)均為正數(shù)的等差數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，則SKIPIF1<0的最小值為__________.9．已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0表示數(shù)列SKIPIF1<0的前n項(xiàng)和，則使不等式SKIPIF1<0成立的正整數(shù)n的最小值是______．三、解答題10．已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且SKIPIF1<0成等比數(shù)列．(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式;(2)設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0，求證:SKIPIF1<0.11．已知數(shù)列SKIPIF1<0的首項(xiàng)SKIPIF1<0，且滿足SKIPIF1<0.(1)求證：數(shù)列SKIPIF1<0為等比數(shù)列；(2)若SKIPIF1<0，求滿足條件的最大整數(shù)SKIPIF1<0.12．近兩年，直播帶貨逐漸成為一種新興的營銷模式，帶來電商行業(yè)的新增長點(diǎn).某直播平臺(tái)第1年初的啟動(dòng)資金為500萬元，由于一些知名主播加入，平臺(tái)資金的年平均增長率可達(dá)SKIPIF1<0，每年年底把除運(yùn)營成本SKIPIF1<0萬元，再將剩余資金繼續(xù)投入直播平合.(1)若SKIPIF1<0，在第3年年底扣除運(yùn)營成本后，直播平臺(tái)的資金有多少萬元？(2)每年的運(yùn)營成本最多控制在多少萬元，才能使得直播平臺(tái)在第6年年底?除運(yùn)營成本后資金達(dá)到3000萬元？（結(jié)果精確到SKIPIF1<0萬元）13．若函數(shù)SKIPIF1<0是其定義域內(nèi)的區(qū)間SKIPIF1<0上的嚴(yán)格增函數(shù)，而SKIPIF1<0是SKIPIF1<0上的嚴(yán)格減函數(shù)，則稱SKIPIF1<0是SKIPIF1<0上的“弱增函數(shù)”.若數(shù)列SKIPIF1<0是嚴(yán)格增數(shù)列，而SKIPIF1<0是嚴(yán)格減數(shù)列，則稱SKIPIF1<0是“弱增數(shù)列”.(1)判斷函數(shù)SKIPIF1<0是否為SKIPIF1<0上的“弱增函數(shù)”，并說明理由（其中SKIPIF1<0是自然對(duì)數(shù)的底數(shù)）；(2)已知函數(shù)SKIPIF1<0與函數(shù)SKIPIF1<0的圖像關(guān)于坐標(biāo)原點(diǎn)對(duì)稱，若SKIPIF1<0是SKIPIF1<