新高考數(shù)學(xué)二輪提升練專題19 數(shù)列中常見的求和問題（原卷版）

專題19數(shù)列中常見的求和問題1、（2021年全國新高考Ⅰ卷數(shù)學(xué)試題）某校學(xué)生在研究民間剪紙藝術(shù)時(shí)，發(fā)現(xiàn)剪紙時(shí)經(jīng)常會沿紙的某條對稱軸把紙對折，規(guī)格為SKIPIF1<0的長方形紙，對折1次共可以得到SKIPIF1<0，SKIPIF1<0兩種規(guī)格的圖形，它們的面積之和SKIPIF1<0，對折2次共可以得到SKIPIF1<0，SKIPIF1<0，SKIPIF1<0三種規(guī)格的圖形，它們的面積之和SKIPIF1<0，以此類推，則對折4次共可以得到不同規(guī)格圖形的種數(shù)為______；如果對折SKIPIF1<0次，那么SKIPIF1<0______SKIPIF1<0.2、（2020年全國統(tǒng)一高考數(shù)學(xué)試卷（理科）（新課標(biāo)Ⅰ））設(shè)SKIPIF1<0是公比不為1的等比數(shù)列，SKIPIF1<0為SKIPIF1<0，SKIPIF1<0的等差中項(xiàng)．（1）求SKIPIF1<0的公比；（2）若SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．3、（2020年全國統(tǒng)一高考數(shù)學(xué)試卷（文科）（新課標(biāo)Ⅲ））設(shè)等比數(shù)列{an}滿足SKIPIF1<0，SKIPIF1<0．（1）求{an}的通項(xiàng)公式；（2）記SKIPIF1<0為數(shù)列{log3an}的前n項(xiàng)和．若SKIPIF1<0，求m．4、（2020年全國統(tǒng)一高考數(shù)學(xué)試卷（理科）（新課標(biāo)Ⅲ））設(shè)數(shù)列{an}滿足a1=3，SKIPIF1<0．（1）計(jì)算a2，a3，猜想{an}的通項(xiàng)公式并加以證明；（2）求數(shù)列{2nan}的前n項(xiàng)和Sn．5、【2021年乙卷文科】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<0．6、【2021年新高考1卷】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.．題組一、利用周期性（規(guī)律性求和）1-1、（2022·江蘇宿遷·高三期末）記SKIPIF1<0表示不超過實(shí)數(shù)SKIPIF1<0的最大整數(shù)，記SKIPIF1<0，則SKIPIF1<0的值為（）A．5479 B．5485 C．5475 D．54822、（2022·湖南郴州·高三期末）高斯是德國著名的數(shù)學(xué)家，近代數(shù)學(xué)奠基者之一，享有“數(shù)學(xué)王子”的稱號.設(shè)SKIPIF1<0，用SKIPIF1<0表示不超過x的最大整數(shù)，則SKIPIF1<0稱為高斯函數(shù).已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，則SKIPIF1<0（）A．4950 B．4953 C．4956 D．4959題組二、裂項(xiàng)相消求和2-1、（2022·河北張家口·高三期末）已知SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.2-2、（2022·河北深州市中學(xué)高三期末）已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0．（1）求數(shù)列SKIPIF1<0的通項(xiàng)公式；（2）設(shè)SKIPIF1<0，若數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，證明：SKIPIF1<0．2-3、（2022·河北唐山·高三期末）已知SKIPIF1<0是數(shù)列SKIPIF1<0的前n項(xiàng)和，SKIPIF1<0，且SKIPIF1<0．(1)證明：SKIPIF1<0為常數(shù)列；(2)若SKIPIF1<0，求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0．題組三、分組求和3-1、（2022·山東萊西·高三期末）已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0為等差數(shù)列；數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0；3-2、（2022·山東青島·高三期末）給定數(shù)列SKIPIF1<0，若滿足SKIPIF1<0，對于任意的SKIPIF1<0，都有SKIPIF1<0，則稱SKIPIF1<0為“指數(shù)型數(shù)列”.(1)已知數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，證明：SKIPIF1<0為“指數(shù)型數(shù)列”；(2)若數(shù)列SKIPIF1<0滿足：SKIPIF1<0；（I）判斷SKIPIF1<0是否為“指數(shù)型數(shù)列”，若是給出證明，若不是說明理由；(Ⅱ)若SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.3-3、（2022·山東臨沂·高三期末）設(shè)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且滿足SKIPIF1<0，SKIPIF1<0．(1)求數(shù)列的通SKIPIF1<0項(xiàng)公式：(2)若SKIPIF1<0，求數(shù)列SKIPIF1<0的前n項(xiàng)和SKIPIF1<0．題組四、錯(cuò)位相減4-1、（2022·湖北·恩施土家族苗族高中高三期末）已知SKIPIF1<0是公差為1的等差數(shù)列，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列．（Ⅰ）求SKIPIF1<0的通項(xiàng)公式；（Ⅱ）求數(shù)列SKIPIF1<0的前n項(xiàng)和．4-2、（2020年全國統(tǒng)一高考數(shù)學(xué)試卷（理科）（新課標(biāo)Ⅰ））設(shè)SKIPIF1<0是公比不為1的等比數(shù)列，SKIPIF1<0為SKIPIF1<0，SKIPIF1<0的等差中項(xiàng)．（1）求SKIPIF1<0的公比；（2）若SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和．4-3、（2022·廣東汕尾·高三期末）已知等比數(shù)列SKIPIF1<0滿足SKIPIF1<0是SKIPIF1<0的等差中項(xiàng)．(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)記SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.題組五、奇偶項(xiàng)5-1、（2022·山東煙臺·高三期末）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0．(1)記SKIPIF1<0，證明：數(shù)列SKIPIF1<0為等比數(shù)列，并求SKIPIF1<0的通項(xiàng)公式；(2)求數(shù)列SKIPIF1<0的前2n項(xiàng)和SKIPIF1<0．5-2、(2022·江蘇新高考基地學(xué)校第一次大聯(lián)考期中)(10分)已知等差數(shù)列eq{a\s\do(n)}滿足eqa\s\do(n)＋a\s\do(n＋1)＝4n，n∈N*．(1)求eq{a\s\do(n)}的通項(xiàng)公式；(2)設(shè)eqb\s\do(1)＝1，b\s\do(n＋1)＝\B\lc\{(\a\al(a\s\do(n)，n為奇數(shù)，,－b\s\do(n)＋2\s\up6(n)，n為偶數(shù)，))求數(shù)列eq{b\s\do(n)}的前2n項(xiàng)和eqS\s\do(2n)．5-3、(2022·江蘇南通海安市期中)已知數(shù)列eq{a\s\do(n)}滿足a1＝1，an＋1＝EQ\B\lc\{(\a\al(\l(2a\S\DO(n)，n為奇數(shù)，),\l(a\S\DO(n)＋3，n為偶數(shù)．)))(1)從下面兩個(gè)條件中選一個(gè)，寫出b1，b2，并求數(shù)列eq{b\s\do(n)}的通項(xiàng)公式；①bn＝a2n－1＋3；②bn＝a2n＋1－a2n－1．(2)求數(shù)列eq{a\s\do(n)}的前n項(xiàng)和為Sn．1、（2022·江蘇如東·高三期末）已知數(shù)列{an}的各項(xiàng)均為正數(shù)，其前n頁和為Sn，且a1＝2，SKIPIF1<0.(1)證明：數(shù)列SKIPIF1<0是等比數(shù)列；(2)求數(shù)列SKIPIF1<0的前n項(xiàng)和.2、（2022·廣東汕尾·高三期末）已知等比數(shù)列SKIPIF1<0滿足SKIPIF1<0是SKIPIF1<0的等差中項(xiàng)．(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式；(2)記SKIPIF1<0，求數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0.3、．(2022·江蘇南通市區(qū)期中)(本題滿分10分)已知數(shù)列eq{a\s\do(n)}是公比為正數(shù)的等比數(shù)列，且eqa\s\do(1)＝2，a\s\do(3)＝a\s\do(2)＋4．(1)求數(shù)列eq{a\s\do(n)}的通項(xiàng)公式；(2)若eqb\s\do(n)＝log\s\do(2)a\s\do(n)，求數(shù)列eq{a\s\do(n)＋b\s\do(n)}的前n項(xiàng)和Sn．4、（2022·廣東·鐵一中學(xué)高三期末）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0