第6節(jié) 數(shù)列中的奇偶項(xiàng)、子數(shù)列問題

第6節(jié)數(shù)列中的奇偶項(xiàng)、子數(shù)列問題考試要求1.明確數(shù)列奇偶項(xiàng)問題的類型，掌握其解決方法.2.會用化歸的思想方法求解子數(shù)列問題．考點(diǎn)一奇偶項(xiàng)問題角度1含有(－1)n的類型例1已知bn＝(－1)nn2，求數(shù)列{bn}的前n項(xiàng)和Tn.____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________角度2已知條件明確的奇偶項(xiàng)問題例2(2023·新高考Ⅱ卷節(jié)選)已知an＝2n＋3，bn＝eq\b\lc\{(\a\vs4\al\co1(an－6，n是奇數(shù)，,2an，n是偶數(shù).))記Sn，Tn分別為數(shù)列{an}，{bn}的前n項(xiàng)和，證明：當(dāng)n>5時(shí)，Tn>Sn.____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________感悟提升1.含有(－1)n的數(shù)列求和問題一般采用分組(并項(xiàng))法求和；2．對于通項(xiàng)公式奇、偶項(xiàng)不同的數(shù)列{an}求Sn時(shí)，我們可以分別求出奇數(shù)項(xiàng)的和與偶數(shù)項(xiàng)的和，也可以先求出S2k，再利用S2k－1＝S2k－a2k，求S2k－1.訓(xùn)練1(2024·青島模擬)已知遞增等比數(shù)列{an}的前n項(xiàng)和為Sn，且S3＝13，aeq\o\al(2,3)＝3a4，等差數(shù)列{bn}滿足b1＝a1，b2＝a2－1.(1)求數(shù)列{an}和{bn}的通項(xiàng)公式；(2)若cn＝eq\b\lc\{(\a\vs4\al\co1(－an＋1bn，n為奇數(shù)，,anbn，n為偶數(shù)，))請判斷c2n－1＋c2n與a2n的大小關(guān)系，并求數(shù)列{cn}的前20項(xiàng)和．____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________考點(diǎn)二子數(shù)列問題角度1公共項(xiàng)例3已知an＝3n－1，bn＝3n＋n，n∈N*.若數(shù)列{an}與{bn}中有公共項(xiàng)，即存在k，m∈N*，使得ak＝bm成立．按照從小到大的順序?qū)⑦@些公共項(xiàng)排列，得到一個(gè)新的數(shù)列，記作{cn}，求c1＋c2＋…＋cn.____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________角度2插項(xiàng)、提項(xiàng)例4已知數(shù)列{an}中，an＝eq\b\lc\{(\a\vs4\al\co1(3，n＝1，,2n－1，n≥2，))若保持{an}中各項(xiàng)先后順序不變，在ak與ak＋1之間插入k個(gè)1，使它們和原數(shù)列的項(xiàng)構(gòu)成一個(gè)新的數(shù)列{bn}，記{bn}的前n項(xiàng)和為Tn，求T100的值(用數(shù)字作答)．____________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________________感悟提升1.兩個(gè)等差(比)數(shù)列的公共項(xiàng)是等差(比)數(shù)列，且公差(比)是兩等差(比)數(shù)列公差(比)的最小公倍數(shù)，一個(gè)等差與一個(gè)等比數(shù)列的公共項(xiàng)，則要通過其項(xiàng)數(shù)之間的關(guān)系來確定．2．?dāng)?shù)列的插項(xiàng)、提項(xiàng)問題可通過研究前n次的變化探究出一般性規(guī)律，從而確定新數(shù)列的首項(xiàng)、項(xiàng)數(shù)、公差(或公比)、末項(xiàng)等信息．訓(xùn)練2(2024·濟(jì)南模擬)已知等差數(shù)列{an}和等比數(shù)列{bn}滿足a1＝4，b1＝2，a2＝2b2－1，a3＝b3＋2.(1)求{an}和{bn}的通項(xiàng)公式；(2)數(shù)列{an}和{bn}中的所有項(xiàng)分別構(gòu)成集合A，B，將A∪B的所有元素按從小到大依次排列構(gòu)成一個(gè)新數(shù)列{cn}，求數(shù)列{cn}的前60項(xiàng)和S60._______________________________________________________________________________________________________________________________________________________________________