上傳人：M***

認(rèn)證信息

認(rèn)證類(lèi)型：個(gè)人認(rèn)證

認(rèn)證主體：楊**（實(shí)名認(rèn)證）

IP屬地：廣西

IP屬地：廣西 上傳時(shí)間：2024-12-14 格式：DOC 頁(yè)數(shù)：8 大小：395KB 積分：6 舉報(bào) 版權(quán)申訴

答案第=page11頁(yè)，共=sectionpages22頁(yè)專(zhuān)題14數(shù)列的通項(xiàng)公式?？记蠓ā揪毣A(chǔ)】單選題1．（2023·四川成都·統(tǒng)考一模）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0．若SKIPIF1<0，則SKIPIF1<0（

）A．512 B．510 C．256 D．2542．（2023·四川攀枝花·統(tǒng)考二模）已知正項(xiàng)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，設(shè)SKIPIF1<0，數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，則滿(mǎn)足SKIPIF1<0的n的最小正整數(shù)解為（

）A．15 B．16 C．3 D．43．（2022·陜西咸陽(yáng)·武功縣普集高級(jí)中學(xué)統(tǒng)考模擬預(yù)測(cè)）南宋數(shù)學(xué)家楊輝所著的《詳解九章算法商功》中出現(xiàn)了如圖所示的形狀后人稱(chēng)為“三角垛”（如圖所示的是一個(gè)4層的三角躁）,“三角垛”最上層有1個(gè)球,第二層有3個(gè)球,第三層有6個(gè)球,…,設(shè)第SKIPIF1<0層有SKIPIF1<0個(gè)球,從上往下SKIPIF1<0層球的總數(shù)為SKIPIF1<0,則下列結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2022·河南開(kāi)封·統(tǒng)考一模）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和SKIPIF1<0，若SKIPIF1<0，則SKIPIF1<0（

）A．8 B．16 C．32 D．645．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，則SKIPIF1<0（

）A．0 B．50 C．100 D．25256．（2023·四川內(nèi)江·統(tǒng)考一模）已知數(shù)列SKIPIF1<0滿(mǎn)足：SKIPIF1<0，點(diǎn)SKIPIF1<0在函數(shù)SKIPIF1<0的圖象上，記SKIPIF1<0為SKIPIF1<0的前n項(xiàng)和，則SKIPIF1<0（

）A．3 B．4 C．5 D．67．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知數(shù)列{SKIPIF1<0}滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則數(shù)列{SKIPIF1<0}第2022項(xiàng)為（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<08．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知正項(xiàng)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，SKIPIF1<0，記SKIPIF1<0，若數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．200 D．4009．（2023·黑龍江·黑龍江實(shí)驗(yàn)中學(xué)?？家荒＃┮阎猄KIPIF1<0是等比數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和，且SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．SKIPIF1<010．（2022秋·山西·高三統(tǒng)考期中）已知數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，則下列正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<011．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知SKIPIF1<0為數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)之和，且滿(mǎn)足SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0為等差數(shù)列 B．若SKIPIF1<0為等差數(shù)列，則公差為2C．SKIPIF1<0可能為等比數(shù)列 D．SKIPIF1<0的最小值為0，最大值為2012．（2022秋·黑龍江綏化·高三?？茧A段練習(xí)）南宋數(shù)學(xué)家楊輝所著的《詳解九章算法·商功》中出現(xiàn)了如圖所示的形狀，后人稱(chēng)為“三角垛”（下圖所示的是一個(gè)4層的三角跺）.“三角垛”最上層有1個(gè)球，第二層有3個(gè)球，第三層有6個(gè)球，…，設(shè)第n層有SKIPIF1<0個(gè)球，從上往下n層球的球的總數(shù)為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0二、多選題13．（2023·全國(guó)·校聯(lián)考模擬預(yù)測(cè)）記函數(shù)SKIPIF1<0在SKIPIF1<0處的導(dǎo)數(shù)為SKIPIF1<0，則SKIPIF1<0________．14．（2023秋·江蘇揚(yáng)州·高三?？计谀┮阎獢?shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，若數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0_____________.15．（2023·江蘇蘇州·蘇州中學(xué)?？寄M預(yù)測(cè)）數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0__________16．（2022·上海青浦·統(tǒng)考一模）已知數(shù)列SKIPIF1<0中，SKIPIF1<0，記SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，且滿(mǎn)足SKIPIF1<0．若對(duì)任意SKIPIF1<0，都有SKIPIF1<0，則首項(xiàng)SKIPIF1<0的取值范圍是______．三、填空題17．（2023·云南紅河·統(tǒng)考一模）已知正項(xiàng)數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且滿(mǎn)足SKIPIF1<0．(1)求數(shù)列SKIPIF1<0的通項(xiàng)公式：(2)若SKIPIF1<0，數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，證明：SKIPIF1<0．四、解答題18．（2023·河南鄭州·統(tǒng)考一模）已知數(shù)列SKIPIF1<0滿(mǎn)足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前SKIPIF1<0項(xiàng)和SKIPIF1<0．【提能力】一、單選題19．（2022·浙江·統(tǒng)考高考真題）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<020．（2021·浙江·統(tǒng)考高考真題）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0.記數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<021．（2018·陜西安康·統(tǒng)考三模）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．2525 C．SKIPIF1<0 D．252622．（2022秋·山東濰坊·高三?？茧A段練習(xí)）已知數(shù)列SKIPIF1<0和SKIPIF1<0首項(xiàng)均為1，且SKIPIF1<0，SKIPIF1<0，數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且滿(mǎn)足SKIPIF1<0，則SKIPIF1<0（

）A．2019 B．SKIPIF1<0 C．4037 D．SKIPIF1<023．（2022秋·江蘇鎮(zhèn)江·高三揚(yáng)中市第二高級(jí)中學(xué)?？计谀?022年第二十四屆北京冬奧會(huì)開(kāi)幕式上由96片小雪花組成的大雪花驚艷了全世界，數(shù)學(xué)中也有一朵美麗的雪花一“科赫雪花”．它可以這樣畫(huà)，任意畫(huà)一個(gè)正三角形SKIPIF1<0，并把每一邊三等分：取三等分后的一邊中間一段為邊向外作正三角形，并把這“中間一段”擦掉，形成雪花曲線SKIPIF1<0；重復(fù)上述兩步，畫(huà)出更小的三角形．一直重復(fù)，直到無(wú)窮，形成雪花曲線，SKIPIF1<0．設(shè)雪花曲線SKIPIF1<0的邊長(zhǎng)為SKIPIF1<0，邊數(shù)為SKIPIF1<0，周長(zhǎng)為SKIPIF1<0，面積為SKIPIF1<0，若SKIPIF1<0，則下列說(shuō)法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0均構(gòu)成等比數(shù)列 D．SKIPIF1<024．（2022·全國(guó)·高三專(zhuān)題練習(xí)）“中國(guó)剩余定理”又稱(chēng)“孫子定理”，最早可見(jiàn)于我國(guó)南北朝時(shí)期的數(shù)學(xué)著作《孫子算經(jīng)》．1852年，英國(guó)傳教士偉烈亞力將該解法傳至歐洲，1874年，英國(guó)數(shù)學(xué)家馬西森指出此法符合1801年由高斯得到的關(guān)于同余式解法的一般性定理，因而西方稱(chēng)之為“中國(guó)剩余定理”．此定理講的是關(guān)于整除的問(wèn)題，現(xiàn)將SKIPIF1<0到SKIPIF1<0這SKIPIF1<0個(gè)數(shù)中，能被SKIPIF1<0除余SKIPIF1<0且被SKIPIF1<0除余SKIPIF1<0的數(shù)按從小到大的順序排成一列，構(gòu)成數(shù)列SKIPIF1<0，則該數(shù)列共有（

）A．SKIPIF1<0項(xiàng) B．SKIPIF1<0項(xiàng) C．SKIPIF1<0項(xiàng) D．SKIPIF1<0項(xiàng)25．（2022·全國(guó)·高三專(zhuān)題練習(xí)）數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則下列結(jié)論錯(cuò)誤的是（

）A．SKIPIF1<0 B．SKIPIF1<0是等比數(shù)列C．SKIPIF1<0 D．SKIPIF1<026．（2023·全國(guó)·高三專(zhuān)題練習(xí)）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0表示不超過(guò)x的最大整數(shù)（例如SKIPIF1<0，SKIPIF1<0）.則SKIPIF1<0（

）A．2018 B．2019 C．2020 D．2021二、多選題27．（2023·全國(guó)·高三專(zhuān)題練習(xí)）數(shù)列SKIPIF1<0首項(xiàng)SKIPIF1<0，對(duì)一切正整數(shù)SKIPIF1<0，都有SKIPIF1<0，則（

）A．對(duì)一切正整數(shù)SKIPIF1<0都有SKIPIF1<0 B．?dāng)?shù)列SKIPIF1<0單調(diào)遞減C．存在正整數(shù)SKIPIF1<0，使得SKIPIF1<0 D．SKIPIF1<0都是數(shù)列SKIPIF1<0的項(xiàng)28．（2022秋·湖南岳陽(yáng)·高三?？茧A段練習(xí)）設(shè)首項(xiàng)為1的數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，若SKIPIF1<0SKIPIF1<0，則下列結(jié)論正確的是（）A．?dāng)?shù)列SKIPIF1<0為等比數(shù)列B．?dāng)?shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0C．?dāng)?shù)列SKIPIF1<0為等比數(shù)列D．?dāng)?shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<029．（2021·全國(guó)·高三專(zhuān)題練習(xí)）斐波那契螺旋線，也稱(chēng)“黃金螺旋”，是根據(jù)斐波那契數(shù)列畫(huà)出來(lái)的螺旋曲線，自然界中存在許多斐波那契螺旋線的圖案，是自然界最完美的經(jīng)典黃金比例.作圖規(guī)則是在以斐波那契數(shù)為邊的正方形拼成的長(zhǎng)方形，然后在正方形里面畫(huà)一個(gè)90度的扇形，連起來(lái)的弧線就是斐波那契螺旋線.它來(lái)源于斐波那契數(shù)列，又稱(chēng)為黃金分割數(shù)列.現(xiàn)將斐波那契數(shù)列記為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，邊長(zhǎng)為斐波那契數(shù)SKIPIF1<0的正方形所對(duì)應(yīng)扇形面積記為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<030．（2022·全國(guó)·高三專(zhuān)題練習(xí)）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，其前SKIPIF1<0項(xiàng)和為SKIPIF1<0，則下列結(jié)論中正確的有（

）A．SKIPIF1<0是遞增數(shù)列 B．SKIPIF1<0是等比數(shù)列C．SKIPIF1<0 D．SKIPIF1<0三、填空題31．（2022·陜西西安·西安中學(xué)?？寄M預(yù)測(cè)）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，SKIPIF1<0，則數(shù)列SKIPIF1<0的前100項(xiàng)和SKIPIF1<0______．32．（2023·陜西寶雞·校聯(lián)考模擬預(yù)測(cè)）已知數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0，則數(shù)列SKIPIF1<0的前2022項(xiàng)的和為_(kāi)__________.33．（2022春·河南鄭州·高三校聯(lián)考階段練習(xí)）數(shù)列SKIPIF1<0滿(mǎn)足SKIPIF1<0(SKIPIF1<0，且SKIPIF1<0)，SKIPIF1<0，對(duì)于任意SKIPIF1<0有SKIPIF1<0恒成立，則SKIPIF1<0的取值范圍是___________.34．（2022秋·河南南陽(yáng)·高三南陽(yáng)中學(xué)?？茧A段練習(xí)）已知數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，且SKIPIF1<0，設(shè)函數(shù)SKIPIF1<0，