新高考數(shù)學(xué)二輪復(fù)習(xí) 小題綜合練專題08 數(shù)列（原卷版）

專題08數(shù)列小題綜合一、單選題1．（2023·浙江寧波·鎮(zhèn)海中學(xué)?？寄M預(yù)測(cè)）數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．32．（2023·浙江溫州·樂(lè)清市知臨中學(xué)校考模擬預(yù)測(cè)）已知等比數(shù)列SKIPIF1<0的前n項(xiàng)和為SKIPIF1<0，公比為q，且SKIPIF1<0，則（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<03．（2023·浙江嘉興·統(tǒng)考模擬預(yù)測(cè)）數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為SKIPIF1<0，則數(shù)列SKIPIF1<0的前SKIPIF1<0項(xiàng)和為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<04．（2023·浙江·高三專題練習(xí)）已知公差不為零的等差數(shù)列SKIPIF1<0滿足：SKIPIF1<0，且SKIPIF1<0成等比數(shù)列，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<05．（2023春·浙江杭州·高三浙江省杭州第二中學(xué)校聯(lián)考階段練習(xí)）已知等差數(shù)列SKIPIF1<0的公差為d，前n項(xiàng)和為SKIPIF1<0，則“SKIPIF1<0”是“SKIPIF1<0”的（

）A．充分不必要條件 B．必要不充分條件C．充分必要條件 D．既不充分也不必要條件6．（2023·浙江·高三專題練習(xí)）已知SKIPIF1<0是公差不為0的等差數(shù)列，SKIPIF1<0，若SKIPIF1<0成等比數(shù)列，則SKIPIF1<0（

）A．2023 B．2024 C．4046 D．40487．（2023·浙江溫州·統(tǒng)考三模）已知數(shù)列SKIPIF1<0各項(xiàng)為正數(shù)，SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則（

）A．SKIPIF1<0是等差數(shù)列 B．SKIPIF1<0是等比數(shù)列C．SKIPIF1<0是等差數(shù)列 D．SKIPIF1<0是等比數(shù)列8．（2023·浙江溫州·樂(lè)清市知臨中學(xué)?？级＃皸钶x三角”是中國(guó)古代重要的數(shù)學(xué)成就，如圖是由“楊輝三角”拓展而成的三角形數(shù)陣，記SKIPIF1<0為圖中虛線上的數(shù)SKIPIF1<0構(gòu)成的數(shù)列SKIPIF1<0的第SKIPIF1<0項(xiàng)，則SKIPIF1<0的值為（

）

A．1275 B．1276 C．1270 D．12809．（2023春·浙江寧波·高三校聯(lián)考階段練習(xí)）非零實(shí)數(shù)SKIPIF1<0滿足SKIPIF1<0成等差數(shù)列，則SKIPIF1<0的最小值為（

）A．SKIPIF1<0 B．SKIPIF1<0 C．3 D．SKIPIF1<010．（2023春·浙江杭州·高三浙江省杭州第二中學(xué)?？茧A段練習(xí)）數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0，則SKIPIF1<0（

）A．SKIPIF1<0 B．SKIPIF1<0 C．SKIPIF1<0 D．SKIPIF1<0二、多選題11．（2023·浙江·高三專題練習(xí)）“冰雹猜想”也稱為“角谷猜想”，是指對(duì)于任意一個(gè)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(guò)若干次操作后的結(jié)果必為1，猶如冰雹掉落的過(guò)程.參照“冰雹猜想”，提出了如下問(wèn)題：設(shè)SKIPIF1<0，各項(xiàng)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(shí)，SKIPIF1<0B．當(dāng)SKIPIF1<0時(shí)，SKIPIF1<0C．當(dāng)SKIPIF1<0為奇數(shù)時(shí)，SKIPIF1<0D．當(dāng)SKIPIF1<0為偶數(shù)時(shí)，SKIPIF1<0是遞增數(shù)列12．（2023·浙江寧波·鎮(zhèn)海中學(xué)?？寄M預(yù)測(cè)）定義：若數(shù)列SKIPIF1<0滿足，存在實(shí)數(shù)M，對(duì)任意SKIPIF1<0，都有SKIPIF1<0，則稱M是數(shù)列SKIPIF1<0的一個(gè)上界．現(xiàn)已知SKIPIF1<0為正項(xiàng)遞增數(shù)列，SKIPIF1<0，下列說(shuō)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(wú)上界，則對(duì)于任意的SKIPIF1<0，均存在SKIPIF1<0，使得SKIPIF1<0D．若SKIPIF1<0無(wú)上界，則存在SKIPIF1<0，當(dāng)SKIPIF1<0時(shí)，恒有SKIPIF1<013．（2023·浙江·校聯(lián)考三模）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(xiàng)之差并不相等，但是逐項(xiàng)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(xiàng)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(xiàng)分別為5，9，17，27，37，45，49，設(shè)通項(xiàng)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(xiàng)和最大C．SKIPIF1<0D．SKIPIF1<014．（2023·浙江·統(tǒng)考二模）已知等差數(shù)列SKIPIF1<0的公差為d，前n項(xiàng)和是SKIPIF1<0，滿足SKIPIF1<0，則（

）．A．SKIPIF1<0的最小值為SKIPIF1<0 B．SKIPIF1<0C．滿足SKIPIF1<0的n的最大值為4 D．SKIPIF1<015．（2023·浙江·高三專題練習(xí)）定義：若存在正實(shí)數(shù)M使SKIPIF1<0，則稱正數(shù)列SKIPIF1<0為有界正數(shù)列．已知數(shù)列SKIPIF1<0滿足SKIPIF1<0，SKIPIF1<0為數(shù)列SKIPIF1<0的前n項(xiàng)和．則（

）A．?dāng)?shù)列SKIPIF1<0為遞增數(shù)列 B．?dāng)?shù)列SKIPIF1<0為遞增數(shù)列C．?dāng)?shù)列SKIPIF1<0為有界正數(shù)列 D．?dāng)?shù)列SKIPIF1<0為有界正數(shù)列16．（2023·浙江·校聯(lián)考二模）已知遞增數(shù)列SKIPIF1<0的各項(xiàng)均為正整數(shù)，且其前SKIPIF1<0項(xiàng)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<017．（2023·浙江金華·浙江金華第一中學(xué)校考模擬預(yù)測(cè)）已知各項(xiàng)均為正數(shù)的數(shù)列SKIPIF1<0滿足SKIPIF1<0為其前SKIPIF1<0項(xiàng)和，則（

）A．SKIPIF1<0 B．SKIPIF1<0C．SKIPIF1<0 D．SKIPIF1<0三、填空題18．（2023秋·浙江紹興·高三期末）設(shè)SKIPIF1<0是首項(xiàng)為1的數(shù)列，且SKIPIF1<0，則SKIPIF1<0___________．19．（2023·浙江·二模）已知等比數(shù)列SKIPIF1<0滿足SKIPIF1<0，則公比SKIPIF1<0______.20．（2023·浙江嘉興·?？寄M預(yù)測(cè)）已知數(shù)列SKIPIF1<0的通項(xiàng)公式為SKIPIF1<0，數(shù)列SKIPIF1<0是以1為首項(xiàng)，2為公比的等比數(shù)列，則SKIPIF1<0___________.21．（2023·浙江·校聯(lián)考模擬預(yù)測(cè)）定義：對(duì)于數(shù)列SKIPIF1<0，如果存在常數(shù)SKIPIF1<0，使得對(duì)于任意SKIPIF1<0，都有SKIPIF1<0，成立，則稱數(shù)列SKIPIF1<0為“SKIPIF1<0擺動(dòng)數(shù)列”，SKIPIF1<0稱為數(shù)列SKIPIF1<0的擺動(dòng)值.若SKIPIF1<0，且數(shù)列SKIPIF1<0的擺動(dòng)值為0，則SKIPIF1<0的取值范圍為__________.22．（2023·浙江·高三專題練習(xí)）已知數(shù)列SKIPIF1<0，其中第一項(xiàng)是SKIPIF1<0，接下來(lái)的兩項(xiàng)是SKIPIF1<0，再接下來(lái)的三項(xiàng)是SKIPIF1<0，依此類推.將該數(shù)列前SKIPIF1<0項(xiàng)的和記為SKIPIF1<0，則使得SKIPIF1<0成立的最小正整數(shù)SKIPIF1<0的值是______.23．（2023·浙江金華·統(tǒng)考模擬預(yù)測(cè)）數(shù)學(xué)王子高斯在小時(shí)候計(jì)算SKIPIF1<0時(shí)，他是這樣計(jì)算的：