新高考數(shù)學(xué)二輪復(fù)習(xí) 多選題分類提升練習(xí)專題一【數(shù)列】多選題專練六十題原卷版

多選題專練六十題專題一數(shù)列（學(xué)生版）第一部——高考真題練1．（2021·全國·統(tǒng)考高考真題）設(shè)正整數(shù)SKIPIF1<0，其中SKIPIF1<0，記SKIPIF1<0．則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0第二部——基礎(chǔ)模擬題2．（2023·河北張家口·統(tǒng)考三模）已知SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項和，SKIPIF1<0，則下列遞推關(guān)系中能使SKIPIF1<0存在最大值的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<03．（2023·浙江·校聯(lián)考模擬預(yù)測）意大利著名數(shù)學(xué)家萊昂納多．斐波那契（LeonardoFibonacci）在研究兔子繁殖問題時，發(fā)現(xiàn)有這樣一列數(shù)：1，1，2，3，5，8，13，21，34，…，該數(shù)列的特點是：前兩個數(shù)都是1，從第三個數(shù)起，每一個數(shù)都等于它的前面兩個數(shù)的和，人們把這樣的一列數(shù)稱為“斐波那契數(shù)列”．同時，隨著SKIPIF1<0趨于無窮大，其前一項與后一項的比值越來越逼近黃金分割SKIPIF1<0，因此又稱“黃金分割數(shù)列”，記斐波那契數(shù)列為SKIPIF1<0，則下列結(jié)論正確的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<04．（2023·江蘇南京·南京市第一中學(xué)?？寄M預(yù)測）在一次《數(shù)列》的公開課時，有位教師引導(dǎo)學(xué)生構(gòu)造新數(shù)列：在數(shù)列的每相鄰兩項之間插入此兩項的和，形成新的數(shù)列，再把所得數(shù)列按照此方法不斷構(gòu)造出新的數(shù)列.下面我們將數(shù)列1，2進(jìn)行構(gòu)造，第1次得到數(shù)列SKIPIF1<0；第2次得到數(shù)列SKIPIF1<0；第SKIPIF1<0次得到數(shù)列SKIPIF1<0記SKIPIF1<0，數(shù)列SKIPIF1<0的前SKIPIF1<0項為SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<05．（2023·廣東佛山·?？寄M預(yù)測）已知數(shù)列SKIPIF1<0，下列結(jié)論正確的有（

）A．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0B．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0C．若SKIPIF1<0，則數(shù)列SKIPIF1<0是等比數(shù)列D．若SKIPIF1<0為等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和，則數(shù)列SKIPIF1<0為等差數(shù)列6．（2023·安徽亳州·蒙城第一中學(xué)校聯(lián)考模擬預(yù)測）如圖，楊輝三角形中的對角線之和1，1，2，3，5，8，13，21，…構(gòu)成的斐波那契數(shù)列經(jīng)常在自然中神奇地出現(xiàn)，例如向日葵花序中央的管狀花和種子從圓心向外，每一圈的數(shù)字就組成這個數(shù)列，等等.在量子力學(xué)中，粒子糾纏態(tài)、量子臨界點研究也離不開這個數(shù)列.斐波那契數(shù)列SKIPIF1<0的第一項和第二項都是1，第三項起每一項都等于它前兩項的和，則（

）

A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<07．（2023·湖北武漢·武漢二中校聯(lián)考模擬預(yù)測）在公差不為零的等差數(shù)列SKIPIF1<0中，已知其前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0等比數(shù)列，則下列結(jié)論正確的是（

）A．SKIPIF1<0B．SKIPIF1<0C．SKIPIF1<0D．設(shè)數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，則SKIPIF1<08．（2023·江蘇無錫·輔仁高中?？寄M預(yù)測）南宋數(shù)學(xué)家楊輝在《詳解九章算法》和《算法通變本末》中提出了一些新的垛積公式，所討論的高階等差數(shù)列與一般等差數(shù)列不同，前后兩項之差并不相等，但是逐項差數(shù)之差或者高次差成等差數(shù)列.如數(shù)列1，3，6，10，它的前后兩項之差組成新數(shù)列2，3，4，新數(shù)列2，3，4為等差數(shù)列，則數(shù)列1，3，6，10被稱為二階等差數(shù)列，現(xiàn)有高階等差數(shù)列SKIPIF1<0?其前7項分別為5，9，17，27，37，45，49，設(shè)通項公式SKIPIF1<0.則下列結(jié)論中正確的是（

）（參考公式：SKIPIF1<0）A．?dāng)?shù)列SKIPIF1<0為二階等差數(shù)列B．?dāng)?shù)列SKIPIF1<0的前11項和最大C．SKIPIF1<0D．SKIPIF1<09．（2023·遼寧沈陽·東北育才學(xué)校?？寄M預(yù)測）已知數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，且滿足SKIPIF1<0，SKIPIF1<0，則下列說法正確的是（

）A．?dāng)?shù)列SKIPIF1<0的前n項和為SKIPIF1<0B．?dāng)?shù)列SKIPIF1<0的通項公式為SKIPIF1<0C．?dāng)?shù)列SKIPIF1<0不是遞增數(shù)列D．?dāng)?shù)列SKIPIF1<0為遞增數(shù)列10．（2023·湖北武漢·湖北省武昌實驗中學(xué)校考模擬預(yù)測）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0（SKIPIF1<0，SKIPIF1<0為常數(shù)），則下列結(jié)論正確的有（

）A．SKIPIF1<0一定是等比數(shù)列 B．當(dāng)SKIPIF1<0時，SKIPIF1<0C．當(dāng)SKIPIF1<0時，SKIPIF1<0 D．SKIPIF1<011．（2023·湖北省直轄縣級單位·統(tǒng)考模擬預(yù)測）在平面直角坐標(biāo)系SKIPIF1<0中，SKIPIF1<0，B為坐標(biāo)原點，點P在圓SKIPIF1<0上，若對于SKIPIF1<0，存在數(shù)列SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0為公差為2的等差數(shù)列 B．SKIPIF1<0為公比為SKIPIF1<0的等比數(shù)列C．SKIPIF1<0 D．SKIPIF1<0前n項和SKIPIF1<012．（2023·全國·模擬預(yù)測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0為SKIPIF1<0的前SKIPIF1<0項和．則下列說法正確的是（

）A．SKIPIF1<0取最大值時，SKIPIF1<0 B．當(dāng)SKIPIF1<0取最小值時，SKIPIF1<0C．當(dāng)SKIPIF1<0取最大值時，SKIPIF1<0 D．SKIPIF1<0的最大值為SKIPIF1<013．（2023·浙江寧波·鎮(zhèn)海中學(xué)?？寄M預(yù)測）定義：若數(shù)列SKIPIF1<0滿足，存在實數(shù)M，對任意SKIPIF1<0，都有SKIPIF1<0，則稱M是數(shù)列SKIPIF1<0的一個上界．現(xiàn)已知SKIPIF1<0為正項遞增數(shù)列，SKIPIF1<0，下列說法正確的是（

）A．若SKIPIF1<0有上界，則SKIPIF1<0一定存在最小的上界B．若SKIPIF1<0有上界，則SKIPIF1<0可能不存在最小的上界C．若SKIPIF1<0無上界，則對于任意的SKIPIF1<0，均存在SKIPIF1<0，使得SKIPIF1<0D．若SKIPIF1<0無上界，則存在SKIPIF1<0，當(dāng)SKIPIF1<0時，恒有SKIPIF1<014．（2023·廣東佛山·?？寄M預(yù)測）所有的有理數(shù)都可以寫成兩個整數(shù)的比，例如SKIPIF1<0如何表示成兩個整數(shù)的比值呢？SKIPIF1<0代表了等比數(shù)列SKIPIF1<0的無限項求和，可通過計算該數(shù)列的前SKIPIF1<0項的和，再令SKIPIF1<0獲得答案.此時SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0，即可得SKIPIF1<0.則下列說法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0為無限循環(huán)小數(shù)C．SKIPIF1<0為有限小數(shù)D．?dāng)?shù)列SKIPIF1<0的無限項求和是有限小數(shù)15．（2023·山東·山東省實驗中學(xué)?？级＃┢矫媛菪且砸粋€固定點開始，向外圈逐漸旋繞而形成的圖案，如圖（1）．它的畫法是這樣的：正方形ABCD的邊長為4，取正方形ABCD各邊的四等分點E，F(xiàn)，G，H作第二個正方形，然后再取正方形EFGH各邊的四等分點M，N，P，Q作第三個正方形，以此方法一直循環(huán)下去，就可得到陰影部分圖案，設(shè)正方形ABCD邊長為SKIPIF1<0，后續(xù)各正方形邊長依次為SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0，…；如圖（2）陰影部分，設(shè)直角三角形AEH面積為SKIPIF1<0，后續(xù)各直角三角形面積依次為SKIPIF1<0，SKIPIF1<0，…，SKIPIF1<0，…．則（

）

A．?dāng)?shù)列SKIPIF1<0是以4為首項，SKIPIF1<0為公比的等比數(shù)列B．從正方形SKIPIF1<0開始，連續(xù)SKIPIF1<0個正方形的面積之和為32C．使得不等式SKIPIF1<0成立的SKIPIF1<0的最大值為3D．?dāng)?shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<016．（2023·重慶沙坪壩·重慶八中?？级＃┰跀?shù)列SKIPIF1<0中，SKIPIF1<0（SKIPIF1<0，SKIPIF1<0為非零常數(shù)），則稱SKIPIF1<0為“等方差數(shù)列”，SKIPIF1<0稱為“公方差”，下列對“等方差數(shù)列”的判斷正確的是（

）A．SKIPIF1<0是等方差數(shù)列B．若正項等方差數(shù)列SKIPIF1<0的首項SKIPIF1<0，且SKIPIF1<0是等比數(shù)列，則SKIPIF1<0C．等比數(shù)列不可能為等方差數(shù)列D．存在數(shù)列SKIPIF1<0既是等差數(shù)列，又是等方差數(shù)列17．（2023·湖南郴州·校聯(lián)考模擬預(yù)測）已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若數(shù)列SKIPIF1<0和SKIPIF1<0均為等差數(shù)列，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<018．（2022·全國·模擬預(yù)測）已知等比數(shù)列SKIPIF1<0滿足SKIPIF1<0，公比SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，則（）A．SKIPIF1<0 B．當(dāng)SKIPIF1<0時，SKIPIF1<0最小C．當(dāng)SKIPIF1<0時，SKIPIF1<0最小 D．存在SKIPIF1<0，使得SKIPIF1<019．（2023·安徽合肥·合肥市第六中學(xué)?？寄M預(yù)測）已知數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，則（

）A．SKIPIF1<0，使得SKIPIF1<0 B．SKIPIF1<0，使得SKIPIF1<0C．SKIPIF1<0，使得SKIPIF1<0 D．若SKIPIF1<0，則SKIPIF1<020．（2023·江蘇揚(yáng)州·揚(yáng)州中學(xué)?？寄M預(yù)測）設(shè)等比數(shù)列SKIPIF1<0的公比為SKIPIF1<0，其前SKIPIF1<0項和為SKIPIF1<0，前SKIPIF1<0項積為SKIPIF1<0，并且滿足條件SKIPIF1<0，則下列結(jié)論正確的是（）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0的最大值為SKIPIF1<0 D．SKIPIF1<0的最大值為SKIPIF1<021．（2023·湖北武漢·華中師大一附中?？寄M預(yù)測）已知集合SKIPIF1<0，SKIPIF1<0，集合SKIPIF1<0，將集合C中所有元素從小到大依次排列為一個數(shù)列SKIPIF1<0，SKIPIF1<0為數(shù)列SKIPIF1<0的前n項和，則（

）A．SKIPIF1<0B．SKIPIF1<0或2C．SKIPIF1<0D．若存在SKIPIF1<0使SKIPIF1<0，則n的最小值為2622．（2023·浙江·校聯(lián)考模擬預(yù)測）已知函數(shù)SKIPIF1<0的定義域均為SKIPIF1<0.若SKIPIF1<0時SKIPIF1<0，且SKIPIF1<0時SKIPIF1<0，則（

）A．SKIPIF1<0 B．函數(shù)SKIPIF1<0的圖像關(guān)于點SKIPIF1<0對稱C．SKIPIF1<0 D．SKIPIF1<023．（2023·全國·模擬預(yù)測）已知數(shù)列1，1，2，3，5，8，…被稱為“斐波那契數(shù)列”該數(shù)列是以兔子繁殖為例子引入的，故又稱為“兔子數(shù)列”，斐波那契數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<024．（2023·重慶·統(tǒng)考模擬預(yù)測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，記數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若存在正整數(shù)SKIPIF1<0，SKIPIF1<0，使得SKIPIF1<0，則SKIPIF1<0的值是（

）A．1 B．2 C．3 D．425．（2023·江蘇鎮(zhèn)江·江蘇省鎮(zhèn)江第一中學(xué)校考模擬預(yù)測）已知等差數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0B．若SKIPIF1<0，則SKIPIF1<0的最小值為SKIPIF1<0C．SKIPIF1<0取最小值時SKIPIF1<0D．設(shè)SKIPIF1<0，則SKIPIF1<026．（2023·安徽滁州·安徽省定遠(yuǎn)中學(xué)?？寄M預(yù)測）已知首項為SKIPIF1<0，公比為SKIPIF1<0的等比數(shù)列SKIPIF1<0，其前SKIPIF1<0項和為SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列，記SKIPIF1<0，SKIPIF1<0，則（

）A．公比SKIPIF1<0B．若SKIPIF1<0是遞減數(shù)列，則SKIPIF1<0C．若SKIPIF1<0不單調(diào)，則SKIPIF1<0的最大項為SKIPIF1<0D．若SKIPIF1<0不單調(diào)，則SKIPIF1<0的最小項為SKIPIF1<027．（2023·安徽滁州·安徽省定遠(yuǎn)中學(xué)校考模擬預(yù)測）歷史上著名的伯努利錯排問題指的是：一個人有SKIPIF1<0封不同的信，投入SKIPIF1<0個對應(yīng)的不同的信箱，他把每封信都投錯了信箱，投錯的方法數(shù)為SKIPIF1<0例如兩封信都投錯有SKIPIF1<0種方法，三封信都投錯有SKIPIF1<0種方法，通過推理可得：SKIPIF1<0.高等數(shù)學(xué)給出了泰勒公式：SKIPIF1<0，則下列說法正確的是（

）A．SKIPIF1<0B．SKIPIF1<0為等比數(shù)列C．SKIPIF1<0D．信封均被投錯的概率大于SKIPIF1<028．（2023·全國·模擬預(yù)測）設(shè)SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項和．下面幾個條件中，能推出SKIPIF1<0是等差數(shù)列的為（

）A．當(dāng)SKIPIF1<0時，SKIPIF1<0 B．當(dāng)SKIPIF1<0時，SKIPIF1<0C．當(dāng)SKIPIF1<0時，SKIPIF1<0 D．當(dāng)SKIPIF1<0時，SKIPIF1<029．（2023·山東威?！そy(tǒng)考二模）已知數(shù)列SKIPIF1<0的首項SKIPIF1<0，前n項和為SKIPIF1<0．設(shè)SKIPIF1<0與k是常數(shù)，若對任意SKIPIF1<0，均有SKIPIF1<0成立，則稱此數(shù)列為“SKIPIF1<0”數(shù)列．若數(shù)列SKIPIF1<0是“SKIPIF1<0”數(shù)列，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0為等比數(shù)列C．SKIPIF1<0的前n項和為SKIPIF1<0 D．SKIPIF1<0為等差數(shù)列30．（2023·山西陽泉·統(tǒng)考三模）設(shè)無窮數(shù)列SKIPIF1<0為正項等差數(shù)列且其前n項和為SKIPIF1<0，若SKIPIF1<0，則下列判斷正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<031．（2023·湖南岳陽·統(tǒng)考三模）設(shè)數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，且SKIPIF1<0，若SKIPIF1<0，則下列結(jié)論正確的有（

）A．SKIPIF1<0 B．?dāng)?shù)列SKIPIF1<0單調(diào)遞減C．當(dāng)SKIPIF1<0時，SKIPIF1<0取得最小值 D．SKIPIF1<0時，n的最小值為732．（2023·湖南長沙·周南中學(xué)校考三模）已知數(shù)列SKIPIF1<0的前n項和是SKIPIF1<0，則下列說法正確的是（

）A．若SKIPIF1<0，則SKIPIF1<0是等差數(shù)列B．若SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0是等比數(shù)列C．若SKIPIF1<0是等差數(shù)列，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列D．若SKIPIF1<0是等比數(shù)列，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列33．（2023·吉林長春·統(tǒng)考模擬預(yù)測）已知正項數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，且有SKIPIF1<0，則下列結(jié)論正確的是（

）．A．SKIPIF1<0 B．?dāng)?shù)列SKIPIF1<0為等差數(shù)列C．SKIPIF1<0 D．SKIPIF1<034．（2023·江蘇淮安·江蘇省盱眙中學(xué)校考模擬預(yù)測）設(shè)a，SKIPIF1<0，數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列說法不正確的是（

）A．當(dāng)SKIPIF1<0時，SKIPIF1<0 B．當(dāng)SKIPIF1<0時，SKIPIF1<0C．當(dāng)SKIPIF1<0時，SKIPIF1<0 D．當(dāng)SKIPIF1<0時，SKIPIF1<035．（2023·云南昆明·統(tǒng)考模擬預(yù)測）已知a，b，c為非零實數(shù)，則下列說法一定正確的是（

）A．若a，b，c成等比數(shù)列，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列B．若a，b，c成等差數(shù)列，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等差數(shù)列C．若a2，b2，c2成等比數(shù)列，則a，b，c成等比數(shù)列D．若a，b，c成等差數(shù)列，則SKIPIF1<0，SKIPIF1<0，SKIPIF1<0成等比數(shù)列36．（2023·山東·山東省實驗中學(xué)?？家荒＃┮阎猄KIPIF1<0為等差數(shù)列，前n項和為SKIPIF1<0，SKIPIF1<0，公差SKIPIF1<0，則（

）．A．SKIPIF1<0B．SKIPIF1<0C．當(dāng)SKIPIF1<0或6時，SKIPIF1<0取得最大值為30D．?dāng)?shù)列SKIPIF1<0與數(shù)列SKIPIF1<0共有671項互為相反數(shù)37．（2023·江蘇南通·江蘇省如皋中學(xué)校考模擬預(yù)測）意大利著名數(shù)學(xué)家斐波那契在研究兔子的繁殖問題時，發(fā)現(xiàn)有這樣的一列數(shù)：1，1，2，3，5，8，13，21，…．該數(shù)列的特點如下：前兩個數(shù)均為1，從第三個數(shù)起，每一個數(shù)都等于它前面兩個數(shù)的和．人們把這樣的一列數(shù)組成的數(shù)列SKIPIF1<0稱為斐波那契數(shù)列，現(xiàn)將SKIPIF1<0中的各項除以2所得的余數(shù)按原來的順序構(gòu)成的數(shù)列記為SKIPIF1<0，數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，下列說法正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0 D．SKIPIF1<038．（2023·浙江·統(tǒng)考二模）“冰雹猜想”也稱為“角谷猜想”，是指對于任意一個正整數(shù)SKIPIF1<0，如果SKIPIF1<0是奇數(shù)?乘以3再加1，如果SKIPIF1<0是偶數(shù)就除以2，這樣經(jīng)過若干次操作后的結(jié)果必為1，猶如冰雹掉落的過程.參照“冰雹猜想”，提出了如下問題：設(shè)SKIPIF1<0，各項均為正整數(shù)的數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0則（

）A．當(dāng)SKIPIF1<0時，SKIPIF1<0B．當(dāng)SKIPIF1<0時，SKIPIF1<0C．當(dāng)SKIPIF1<0為奇數(shù)時，SKIPIF1<0D．當(dāng)SKIPIF1<0為偶數(shù)時，SKIPIF1<0是遞增數(shù)列39．（2023·海南·統(tǒng)考模擬預(yù)測）已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0，若SKIPIF1<0恒成立，則實數(shù)λ的值可以為（

）A．－36 B．－54 C．－81 D．－10840．（2023·安徽淮北·統(tǒng)考二模）已知棋盤上標(biāo)有第0，1，2，...，100站，棋子開始時位于第0站，棋手拋擲均勻硬幣走跳棋游戲，若擲出正面，棋子向前跳一站；若擲出反面，棋子向前跳兩站，直到跳到第99站（勝利大本營）或第100站（歡樂大本營）時，游戲結(jié)束.設(shè)棋子跳到第n站的概率為SKIPIF1<0，SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0第三部分能力提升模擬題41．（2023·江蘇揚(yáng)州·統(tǒng)考模擬預(yù)測）定義：在數(shù)列的每相鄰兩項之間插入此兩項的積，形成新的數(shù)列，這樣的操作叫作該數(shù)列的一次“美好成長”．將數(shù)列SKIPIF1<0、SKIPIF1<0進(jìn)行“美好成長”，第一次得到數(shù)列SKIPIF1<0、SKIPIF1<0、SKIPIF1<0；第二次得到數(shù)列SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0；SKIPIF1<0；設(shè)第SKIPIF1<0次“美好成長”后得到的數(shù)列為SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0、SKIPIF1<0，并記SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．?dāng)?shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<042．（2023·福建福州·福建省福州第一中學(xué)校考二模）定義在SKIPIF1<0上的函數(shù)SKIPIF1<0，其導(dǎo)函數(shù)分別為SKIPIF1<0，若SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0是奇函數(shù)B．SKIPIF1<0關(guān)于SKIPIF1<0對稱C．SKIPIF1<0周期為4D．SKIPIF1<043．（2023·湖南·鉛山縣第一中學(xué)校聯(lián)考二模）已知數(shù)列SKIPIF1<0，如果存在常數(shù)SKIPIF1<0，對于任意給定的正數(shù)SKIPIF1<0（無論多小），總存在正整數(shù)SKIPIF1<0，使得SKIPIF1<0時，恒有SKIPIF1<0成立，就稱數(shù)列SKIPIF1<0收斂于SKIPIF1<0（極限為SKIPIF1<0），即數(shù)列SKIPIF1<0為收斂數(shù)列.下列結(jié)論正確的是（

）A．?dāng)?shù)列SKIPIF1<0是一個收斂數(shù)列B．若數(shù)列SKIPIF1<0為收斂數(shù)列，則SKIPIF1<0，使得SKIPIF1<0，都有SKIPIF1<0C．若數(shù)列SKIPIF1<0和SKIPIF1<0為收斂數(shù)列，而數(shù)列SKIPIF1<0一定為收斂數(shù)列D．若數(shù)列SKIPIF1<0和SKIPIF1<0為收斂數(shù)列，則數(shù)列SKIPIF1<0不一定為收斂數(shù)列44．（2024·安徽黃山·屯溪一中?？寄M預(yù)測）已知SKIPIF1<0是數(shù)列SKIPIF1<0的前SKIPIF1<0項和，SKIPIF1<0，則（

）A．SKIPIF1<0B．當(dāng)SKIPIF1<0時，SKIPIF1<0C．當(dāng)SKIPIF1<0時，SKIPIF1<0為等差數(shù)列D．當(dāng)數(shù)列SKIPIF1<0單調(diào)遞增時，SKIPIF1<0的取值范圍是SKIPIF1<045．（2023·黑龍江哈爾濱·哈爾濱三中?？寄M預(yù)測）定義在SKIPIF1<0的函數(shù)SKIPIF1<0滿足SKIPIF1<0，且SKIPIF1<0，SKIPIF1<0都有SKIPIF1<0，若方程SKIPIF1<0的解構(gòu)成單調(diào)遞增數(shù)列SKIPIF1<0，則下列說法中正確的是（

）A．SKIPIF1<0B．若數(shù)列SKIPIF1<0為等差數(shù)列，則公差為6C．若SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<046．（2023·山東日照·三模）設(shè)函數(shù)SKIPIF1<0的定義域為SKIPIF1<0，滿足SKIPIF1<0，且當(dāng)SKIPIF1<0時，SKIPIF1<0，則（

）A．SKIPIF1<0B．若對任意SKIPIF1<0，都有SKIPIF1<0，則SKIPIF1<0的取值范圍是SKIPIF1<0C．若方程SKIPIF1<0恰有三個實數(shù)根，則SKIPIF1<0的取值范圍是SKIPIF1<0D．函數(shù)SKIPIF1<0在區(qū)間SKIPIF1<0上的最大值為SKIPIF1<0，若存在SKIPIF1<0，使得SKIPIF1<0成立，則SKIPIF1<047．（2023·湖北黃岡·浠水縣第一中學(xué)?？既＃┤鬝KIPIF1<0為函數(shù)SKIPIF1<0的導(dǎo)函數(shù)，數(shù)列SKIPIF1<0滿足SKIPIF1<0，則稱SKIPIF1<0為“牛頓數(shù)列”.已知函數(shù)SKIPIF1<0，數(shù)列SKIPIF1<0為“牛頓數(shù)列”，其中SKIPIF1<0，則（

）A．SKIPIF1<0B．?dāng)?shù)列SKIPIF1<0是單調(diào)遞減數(shù)列C．SKIPIF1<0D．關(guān)于SKIPIF1<0的不等式SKIPIF1<0的解有無限個48．（2023·浙江·校聯(lián)考二模）已知遞增數(shù)列SKIPIF1<0的各項均為正整數(shù)，且其前SKIPIF1<0項和為SKIPIF1<0，則（

）A．存在公差為1的等差數(shù)列SKIPIF1<0，使得SKIPIF1<0B．存在公比為2的等比數(shù)列SKIPIF1<0，使得SKIPIF1<0C．若SKIPIF1<0，則SKIPIF1<0D．若SKIPIF1<0，則SKIPIF1<049．（2023·江蘇蘇州·校聯(lián)考三模）若數(shù)列SKIPIF1<0滿足：對任意的SKIPIF1<0，總存在SKIPIF1<0，使SKIPIF1<0，則稱SKIPIF1<0是“SKIPIF1<0數(shù)列”.則下列數(shù)列是“SKIPIF1<0數(shù)列”的有（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<050．（2023·全國·模擬預(yù)測）數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，SKIPIF1<0表示SKIPIF1<0落在區(qū)間SKIPIF1<0的項數(shù)，其中SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<051．（2023·廣東廣州·統(tǒng)考三模）設(shè)定義在R上的函數(shù)SKIPIF1<0與SKIPIF1<0的導(dǎo)函數(shù)分別為SKIPIF1<0和SKIPIF1<0．若SKIPIF1<0，SKIPIF1<0，且SKIPIF1<0為奇函數(shù)，則（

）．A．SKIPIF1<0，SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<052．（2023·安徽阜陽·安徽省臨泉第一中學(xué)校考三模）數(shù)列SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，該數(shù)列為著名的裴波那契數(shù)列，它是自然界的產(chǎn)物揭示了花瓣的數(shù)量、樹木的分叉、植物種子的排列等植物的生長規(guī)律，則下面結(jié)論正確的是（

）A．SKIPIF1<0 B．SKIPIF1<0C．?dāng)?shù)列SKIPIF1<0為等比數(shù)列 D．?dāng)?shù)列SKIPIF1<0為等比數(shù)列53．（2023·廣東茂名·統(tǒng)考二模）已知數(shù)列SKIPIF1<0和SKIPIF1<0滿足：SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，SKIPIF1<0，則下列結(jié)論錯誤的是（

）A．?dāng)?shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列 B．僅有有限項使得SKIPIF1<0C．?dāng)?shù)列SKIPIF1<0是遞增數(shù)列 D．?dāng)?shù)列SKIPIF1<0是遞減數(shù)列54．（2023·湖南長沙·長郡中學(xué)?？寄M預(yù)測）已知函數(shù)SKIPIF1<0定義域為SKIPIF1<0，滿足SKIPIF1<0，當(dāng)SKIPIF1<0時，SKIPIF1<0.若函數(shù)SKIPIF1<0的圖象與函數(shù)SKIPIF1<0的圖象的交點為SKIPIF1<0，（其中SKIPIF1<0表示不超過SKIPIF1<0的最大整數(shù)），則（

）A．SKIPIF1<0是偶函數(shù) B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<055．（2023·遼寧沈陽·統(tǒng)考三模）已知等比數(shù)列SKIPIF1<0首項SKIPIF1<0，公比為q，前n項和為SKIPIF1<0，前n項積為SKIPIF1<0，函數(shù)SKIPIF1<0SKIPIF1<0，若SKIPIF1<0，則下列結(jié)論正確的是（

）A．SKIPIF1<0為單調(diào)遞增的等差數(shù)列B．SKIPIF1<0C．SKIPIF1<0為單調(diào)遞增的等比數(shù)列D．使得SKIPIF1<0成立的n的最大值為656．（2023·湖南長沙·長郡中學(xué)校聯(lián)考模擬預(yù)測）“SKIPIF1<0”表示不大于x的最大整數(shù)，例如：SKIPIF1<0，SKIPIF1<0