2021年高考數學真題和模擬題分類匯編專題15推理與證明【含答案】

專題15推理與證明/r/n一、選擇題部分/r/n1/r/n./r/n(2021?貴州畢節(jié)三模/r/n?文/r/nT/r/n9/r/n．/r/n)/r/n如/r/n圖，有甲、乙、丙三個盤子和放在甲盤子中的四塊大小不相同的餅，按下列規(guī)則把餅從甲盤全部移到乙盤中：/r/n①/r/n每次只能移動一塊餅；/r/n②/r/n較大的餅不能放在較小的餅上面，則最少需要移動的次數為（）/r/nA/r/n．/r/n7/r/n /r/nB/r/n．/r/n8/r/n /r/nC/r/n．/r/n15/r/n /r/nD/r/n．/r/n16/r/nC/r/n．/r/n假設甲盤中有/r/nn/r/n塊餅，從甲盤移動到乙盤至少需要/r/na/r/nn/r/n次，則/r/na/r/n1/r/n＝/r/n1/r/n，/r/n當/r/nn/r/n≥/r/n2/r/n時，可先將較大的餅不動，將剩余的/r/nn/r/n﹣/r/n1/r/n塊餅先移動到丙盤中，至少需要移動/r/na/r/nn/r/n﹣/r/n1/r/n次，/r/n再將最大的餅移動到乙盤，需要移動/r/n1/r/n次，/r/n最后將丙盤中所有的丙移動到乙盤中，至少需要移動/r/na/r/nn/r/n﹣/r/n1/r/n次，/r/n由上可知，/r/na/r/nn/r/n＝/r/n2/r/na/r/nn/r/n﹣/r/n1/r/n+1/r/n，且/r/na/r/n1/r/n＝/r/n1/r/n，/r/n所以/r/na/r/n2/r/n＝/r/n2/r/na/r/n1/r/n+1/r/n＝/r/n3/r/n，/r/na/r/n3/r/n＝/r/n2/r/na/r/n2/r/n+1/r/n＝/r/n7/r/n，/r/na/r/n4/r/n＝/r/n2/r/na/r/n3/r/n+1/r/n＝/r/n15/r/n，/r/n則最少需要移動的次數為/r/n15/r/n次．/r/n2.(2021?貴州畢節(jié)三模?文T/r/n5/r/n．)/r/n“干支紀年法”是中國歷法上自古以來使用的紀年方法，甲、乙、丙、丁、戊、己、庚、辛、壬、癸被稱為“十天干”，子、丑、寅、卯、辰、巳、午、未、申、酉、戌、亥叫做“十二地支”．“天干”以“甲”字開始，“地支”以“子”字開始，兩者按干支順序相配，組成了干支紀年法，其相配順序為：甲子、乙丑、丙寅、…、癸酉，甲戌、乙亥、丙子、…、癸未，甲申、乙酉、丙戌、…、癸巳，…，共得到/r/n60/r/n個組合，稱六十甲子，周而復始，無窮無盡．/r/n2021/r/n年是“干支紀年法”中的辛丑年，那么/r/n2015/r/n年是“干支紀年法”中的（）/r/nA/r/n．甲辰年/r/n /r/nB/r/n．乙巳年/r/n /r/nC/r/n．丙午年/r/n /r/nD/r/n．乙未年/r/nD/r/n．/r/n由題意可知，甲、乙、丙、丁、戊、己、庚、辛、壬、癸被稱為“十天干”，/r/n子、丑、寅、卯、辰、巳、午、未、申、酉、戌、亥叫做“十二地支”，/r/n2021/r/n年是“干支紀年法”中的辛丑年，/r/n則/r/n2020/r/n年為庚子，/r/n2019/r/n年為己亥，/r/n2018/r/n年為戊戌，/r/n2017/r/n年為丁酉，/r/n2016/r/n年為丙申，/r/n2015/r/n年為乙未．/r/n3.(2021?江西九江二模?理T/r/n9/r/n．)/r/n古希臘畢達哥拉斯學派認為數是萬物的本源，因此極為重視/r/n數的理論研究，他們常把數描繪成沙灘上的沙?；蛐∈?，并將它們排列成各種形狀進行研究．形數就是指平面上各種規(guī)則點陣所對應的點數，是畢哥拉斯學派最早研究的重要內容之一．如圖是三角形數和四邊形數的前四個數，若三角形數組成數列/r/n{/r/na/r/nn/r/n}/r/n，四邊形數組成數列/r/n{/r/nb/r/nn/r/n}/r/n，記/r/nc/r/nn/r/n＝/r/n，則數列/r/n{c/r/nn/r/n}/r/n的前/r/n10/r/n項和為（）/r/nA/r/n．/r/n /r/nB/r/n．/r/n /r/nC/r/n．/r/n /r/nD/r/n．/r/nD/r/n．/r/n由題意可得，/r/n，/r/n，/r/n所以/r/n，/r/n設數列/r/n{c/r/nn/r/n}/r/n的前/r/nn/r/n項和為/r/nS/r/nn/r/n，/r/n所以/r/n，/r/n所以/r/n．/r/n4/r/n.(2021/r/n?山東濰坊二模?/r/nT6/r/n．/r/n)/r/n關/r/n于函數/r/nf/r/n（/r/nx/r/n）＝/r/n，其中/r/na/r/n，/r/nb/r/n∈/r/nR/r/n，給出下列四個結論：/r/n甲：/r/n6/r/n是該函數的零點；/r/n乙：/r/n4/r/n是該函數的零點；/r/n丙：該函數的零點之積為/r/n0/r/n；/r/n?。悍匠?r/nf/r/n（/r/nx/r/n）＝/r/n有兩個根．/r/n若上述四個結論中有且只有一個結論錯誤，則該錯誤結論是（）/r/nA/r/n．甲/r/n /r/nB/r/n．乙/r/n /r/nC/r/n．丙/r/n /r/nD/r/n．丁/r/nB/r/n．/r/n當/r/nx/r/n∈/r/n[0/r/n，/r/n2]/r/n時，/r/nf/r/n（/r/nx/r/n）＝/r/n2/r/nx/r/n﹣/r/na/r/n為增函數，/r/n當/r/nx/r/n∈/r/n[2/r/n，/r/n+/r/n∞）時，/r/nf/r/n（/r/nx/r/n）＝/r/nb/r/n﹣/r/nx/r/n為減函數，故/r/n6/r/n和/r/n4/r/n只有一個是函數的零點，/r/n即甲乙中有一個結論錯誤，一個結論正確，而丙、丁均正確．/r/n由兩零點之積為/r/n0/r/n，則必有一個零點為/r/n0/r/n，則/r/nf/r/n（/r/n0/r/n）＝/r/n2/r/n0/r/n﹣/r/na/r/n＝/r/n0/r/n，得/r/na/r/n＝/r/n1/r/n，/r/n若甲正確，則/r/nf/r/n（/r/n6/r/n）＝/r/n0/r/n，即/r/nb/r/n﹣/r/n6/r/n＝/r/n0/r/n，/r/nb/r/n＝/r/n6/r/n，/r/n可得/r/nf/r/n（/r/nx/r/n）＝/r/n，由/r/nf/r/n（/r/nx/r/n）＝/r/n，/r/n可得/r/n或/r/n，解得/r/nx/r/n＝/r/n或/r/nx/r/n＝/r/n，方程/r/nf/r/n（/r/nx/r/n）＝/r/n有兩個根，故丁正確．/r/n故甲正確，乙錯誤．/r/n二、填空題部分/r/n5/r/n.(2021/r/n?山西調研二模?文/r/nT/r/n14/r/n)某/r/n校團委為高三學生籌備十八歲成人禮策劃了三種活動方案，分別記作/r/nA/r/n、/r/nB/r/n、/r/nC/r/n，為使活動開展得更加生動有意義，現(xiàn)隨機調查甲、乙、丙三位同學對三種活動方案的喜歡程度/r/n./r/n甲說：/r/n“/r/n我不喜歡方案/r/nA/r/n，但喜歡的活動方案比乙多/r/n./r/n”/r/n乙說：/r/n“/r/n我不喜歡方案/r/nB./r/n”/r/n丙說：/r/n“/r/n我們三人都喜歡同一種方案/r/n”/r/n./r/n由此可以判斷乙喜歡的活動方案是/r/n______./r/nC/r/n．/r/n從丙的說法中推測乙肯定有喜歡的方案，/r/n

從甲的說法中推測甲喜歡/r/n2/r/n種方案，不喜歡方案/r/nA/r/n，那么可以確定是/r/nB/r/n和/r/nC/r/n，/r/n

再從乙的說法中可知，乙只喜歡一種方案，是方案/r/nC/r/n，/r/n

故/r/nC./r/n

/r/n根據三個人所說內容，可以推斷出乙只喜歡一種方案，又丙說：/r/n“/r/n我們三人都喜歡同一種方案/r/n”/r/n，所以可以判斷乙喜歡的活動方案．/r/n

本題主要考查了簡單的合情推理，考查了學生的邏輯推理能力，是基礎題．/r/n

/r/n6.(2021?山東聊城三模?T/r/n13./r/n)/r/n數列/r/n1/r/n，/r/n1/r/n，/r/n2/r/n，/r/n3/r/n，/r/n5/r/n，/r/n8/r/n，/r/n13/r/n，/r/n21/r/n，/r/n34/r/n，/r/n…/r/n稱為斐波那契數列，是意大利著名數學家斐波那契于/r/n1202/r/n年在他寫的《算盤全書》提出的，該數列的特點是：從第三起，每一項都等于它前面兩項的和．在該數列的前/r/n2021/r/n項中，奇數的個數為/r/n________/r/n．/r/n1348/r/n．/r/n【考點】進行簡單的合情推理/r/n由斐波那契數列的特點知：從第一項起，每/r/n3/r/n個數中前兩個為奇數后一個偶數，/r/n∵/r/n2021/r/n3/r/n的整數部分為/r/n673/r/n，余數為/r/n2/r/n∴/r/n該數列的前/r/n2021/r/n項中共有/r/n673/r/n個偶數，奇數的個數為/r/n2021/r/n-/r/n故/r/n1348/r/n

/r/n【分析】由斐波那契數列的特點經過推理即可求得/r/n．/r/n三、解答題部分/r/n7.(2021?高考全國甲卷?理T18)/r/n已知數列/r/n的各項均為正數，記/r/n為/r/n的前/r/nn/r/n項和，從下面①②③中選取兩個作為條件，證明另外一個成立．/r/n①數列/r/n是等差數列：②數列/r/n是等差數列；③/r/n．/r/n注：若選擇不同的組合分別解答，則按第一個解答計分．/r/n選①②作條件證明③時，可設出/r/n，結合/r/n的關系求出/r/n，利用/r/n是等差數列可證/r/n；/r/n選①③作條件證明②時，根據等差數列的求和公式表示出/r/n，結合等差數列定義可證；/r/n選②③作條件證明①時，設出/r/n，結合/r/n的關系求出/r/n，根據/r/n可求/r/n，然后可證/r/n是等差數列./r/n選①②作條件證明③：/r/n設/r/n，則/r/n，/r/n當/r/n時，/r/n；/r/n當/r/n時，/r/n；/r/n因為/r/n也是等差數列，所以/r/n，解得/r/n；/r/n所以/r/n，所以/r/n./r/n選①③作條件證明②：/r/n因為/r/n，/r/n是等差數列，/r/n所以公差/r/n，/r/n所以/r/n，即/r/n，/r/n因為/r/n，/r/n所以/r/n是等差數列./r/n選②③作條件證明①：/r/n設/r/n，則/r/n，/r/n當/r/n時，/r/n；/r/n當/r/n時，/r/n；/r/n因為/r/n，所以/r/n，解得/r/n或/r/n；/r/n當/r/n時，/r/n，當/r/n時，/r/n滿足等差數列的定義，此/r/n時/r/n為等差數列；/r/n當/r/n時，/r/n，/r/n不合題意，舍去./r/n綜上可知/r/n為等差數列./r/n8.(2021?江蘇鹽城三模?T/r/n18/r/n)/r/n請在/r/n①/r/neqa/r/n\/r/ns/r/n\/r/ndo/r/n(1)/r/n＝/r/n\/r/nr/r/n(,2)/r/n；/r/n②/r/neqa/r/n\/r/ns/r/n\/r/ndo/r/n(1)/r/n＝/r/n2/r/n；/r/n③/r/neqa/r/n\/r/ns/r/n\/r/ndo/r/n(1)/r/n＝/r/n3/r/n這/r/n3/r/n個條件中選擇/r/n1/r/n個條件，補全下面的命題使其成為真命題，并證明這個命題/r/n(/r/n選擇多個條件并分別證明的按前/r/n1/r/n個評分/r/n)/r/n．/r/n已知數列/r/neq/r/n{/r/na/r/n\/r/ns/r/n\/r/ndo/r/n(/r/nn/r/n)}/r/n滿足/r/na/r/nn/r/n＋/r/n1/r/n＝/r/na/r/nn/r/n2/r/n，若，則當/r/nn/r/n≥/r/n2/r/n時，/r/na/r/nn/r/n≥/r/n2/r/nn/r/n恒成立．/r/n【考點】數列的通項公式求解與不等式的證明/r/n選/r/n②/r/n．/r/n證明：由/r/na/r/nn/r/n＋/r/n1/r/n＝/r/na/r/nn/r/n2/r/n，且/r/neqa/r/n\/r/ns/r/n\/r/ndo/r/n(1)/r/n＝/r/n2/r/n，所以/r/na/r/nn/r/n＞/r/n0/r/n，/r/n所以/r/nlg/r/na/r/nn/r/n＋/r/n1/r/n＝/r/nlg/r/na/r/nn/r/n，/r/nlg/r/na/r/nn/r/n＝/r/nEQ/r/n2\/r/nS/r/n\/r/nUP/r/n6(/r/nn/r/n－/r/n1)/r/nlg2/r/n，/r/na/r/nn/r/n＝/r/nEQ/r/n2\/r/nS/r/n\/r/nUP/r/n6(2\/r/nS/r/n\/r/nUP/r/n6(/r/nn/r/n－/r/n1）/r/n，/r/n……/r/n5/r/n分/r/n當/r/nn/r/n≥/r/n2/r/n時，只需證明/r/nEQ/r/n2\/r/nS/r/n\/r/nUP/r/n6(/r/nn/r/n－/r/n1)/r/n≥/r/nn/r/n，/r/n令/r/nb/r/nn/r/n＝/r/nEQ/r/n\/r/nF/r/n(/r/nn/r/n,/r/n2/r/n\/r/nS/r/n\/r/nUP/r/n6/r/n(/r/nn/r/n－/r/n1/r/n）/r/n，則/r/nb/r/nn/r/n＋/r/n1/r/n－/r/nb/r/nn/r/n＝/r/nEQ/r/n\/r/nF/r/n(/r/nn/r/n＋/r/n1,/r/n2/r/n\/r/nS/r/n\/r/nUP/r/n6/r/n(/r/nn/r/n）/r/n－/r/nEQ/r/n\/r/nF/r/n(/r/nn/r/n,/r/n2/r/n\/r/nS/r/n\/r/nUP/r/n6/r/n(/r/nn/r/n－/r/n1/r/n）/r/n＝/r/nEQ/r/n\/r/nF/r/n(1/r/n－/r/nn/r/n,/r/n2/r/n\/r/nS/r/n\/r/nUP/r/n6/r/n(/r/nn/r/n）/r/n＜/r/n0/r/n，/r/n……/r/n10/r/n分/r/n所以/r/nb/r/nn/r/n≤/r/nb/r/n2/r/n＝/r/n1/r/n，所以/r/nEQ/r/n2\/r/nS/r/n\/r/nUP/r/n6(/r/nn/r/n－/r/n1)/r/n≥/r/nn/r/n成立/r/n．/r/n綜上所述，當/r/na/r/n1/r/n＝/r/n2/r/n且/r/nn/r/n≥/r/n2/r/n時，/r/na/r/nn/r/n≥/r/n2/r/nn/r/n成立．/r/n……/r/n12/r/n分/r/n注：選/r/n②/r/n為假命題，不得分，選/r/n③/r/n參照給分．/r/n9.(2021?河南開封三模?理T/r/n17/r/n)/r/n已知數列/r/n{/r/na/r/nn/r/n}/r/n滿足/r/na/r/n1/r/n＝﹣/r/n2/r/n，/r/na/r/nn/r/n+1/r/n＝/r/n2/r/na/r/nn/r/n+4/r/n．/r/n（/r/n1/r/n）求/r/na/r/n2/r/n，/r/na/r/n3/r/n，/r/na/r/n4/r/n；/r/n（/r/n2/r/n）猜想/r/n{/r/na/r/nn/r/n}/r/n的通項公式并加以證明；/r/n（/r/n3/r/n）求數列/r/n{|/r/na/r/nn/r/n|}/r/n的前/r/nn/r/n項和/r/nS/r/nn/r/n．/r/n（/r/n1/r/n）由已知，易得/r/na/r/n2/r/n＝/r/n0/r/n，/r/na/r/n3/r/n＝/r/n4/r/n，/r/na/r/n4/r/n＝/r/n12/r/n．/r/n（/r/n2/r/n）猜想/r/n．/r/n因為/r/na/r/nn/r/n+1/r/n＝/r/n2/r/na/r/nn/r/n+4/r/n，所以/r/na/r/nn/r/n+1/r/n+4/r/n＝/r/n2/r/n（/r/na/r/nn/r/n+4/r/n），/r/n，/r/n則/r/n{/r/na/r/nn/r/n+4}/r/n是以/r/n2/r/n為首項，以/r/n2/r/n為公比的等比數列，/r/n所以/r/n，所以＝/r/n．/r/n（/r/n3/r/n）當/r/nn/r/n＝/r/n1/r/n時，/r/na/r/n1/r/n＝﹣/r/n2/r/n＜/r/n0/r/n，/r/nS/r/n1/r/n＝/r/n|/r/na/r/n1/r/n|/r/n＝/r/n2/r/n；/r/n當/r/nn/r/n≥/r/n2/r/n時，/r/na/r/nn/r/n≥/r/n0/r/n，/r/n所以/r/n＝/r/n，/r/n又/r/nn/r/n＝/r/n1/r/n時滿足上式．/r/n所以，當/r/nn/r/n∈/r/nN/r/n*/r/n時，/r/n．/r/n10.(2021?浙江杭州二模?理T/r/n20/r/n．)已/r/n知數列/r/n{/r/na/r/nn/r/n}/r/n，/r/n{/r/nb/r/nn/r/n}/r/n，滿足/r/na/r/nn/r/n＝/r/n2/r/nn/r/n﹣/r/n2/r/n，/r/nb/r/n2/r/nk/r/n﹣/r/n1/r/n＝/r/na/r/nk/r/n（/r/nk/r/n∈/r/nN/r/n*/r/n），/r/nb/r/n2/r/nk/r/n﹣/r/n1/r/n，/r/nb/r/n2/r/nk/r/n，/r/nb/r/n2/r/nk/r/n+1/r/n成等差數列．/r/n（/r/n1/r/n）證明：/r/n{/r/nb/r/n2/r/nk/r/n}/r/n是等比數列；/r/n（/r/n2/r/n）數列/r/n{c/r/nn/r/n}/r/n滿足/r/nc/r/nn/r/n＝/r/n，記數列/r/n{c/r/nn/r/n}/r/n的前/r/nn/r/n項和為/r/nS/r/nn/r/n，求/r/nS/r/nn/r/n．/r/n證明：（/r/n1/r/n）由數列/r/n{/r/na/r/nn/r/n}/r/n，/r/n{/r/nb/r/nn/r/n}/r/n，滿足/r/na/r/nn/r/n＝/r/n2/r/nn/r/n﹣/r/n2/r/n，/r/nb/r/n2/r/nk/r/n﹣/r/n1/r/n＝/r/na/r/nk/r/n（/r/nk/r/n∈/r/nN/r/n*/r/n），/r/n所以/r/n，/r/n由于/r/nb/r/n2/r/nk/r/n﹣/r/n1/r/n，/r/nb/r/n2/r/nk/r/n，/r/nb/r/n2/r/nk/r/n+1/r/n成等差數列．/r/n故/r/n，/r/n整理得/r/n（常數），/r/n所以數列：/r/n{/r/nb/r/n2/r/nk/r/n}/r/n是以/r/n為首項，公比為/r/n2/r/n的等比數列；/r/n（/r/n2/r/n）由于：/r/n{/r/nb/r/n2/r/nk/r/n}/r/n是以/r/n為首項，公比為/r/n2/r/n的等比數列；/r/n所以/r/n，/r/n則/r/n＝/r/n2/r/nn/r/n﹣/r/n3/r/n，/r/n所以/r/n＝/r/n＝/r/n（/r/nn/r/n≥/r/n1/r/n），/r/n則/r/n+/r/n…/r/n+/r/n，/r/n＝/r/n．/r/n1/r/n1/r/n.(2021/r/n?浙江麗水湖州衢州二模?/r/nT20/r/n．/r/n)/r/n已/r/n知數列/r/n{/r/na/r/nn/r/n}/r/n是各項均為正數的等比數列，若/r/na/r/n1/r/n＝/r/n2/r/n，/r/na/r/n2/r/n+/r/na/r/n3/r/n是/r/na/r/n3/r/n與/r/na/r/n4/r/n的等差中項．數列/r/n{/r/nb/r/nn/r/n}/r/n的前/r/nn/r/n項和為/r/nS/r/nn/r/n，且/r/nS/r/nn/r/n+/r/n＝/r/n2/r/na/r/nn/r/n﹣/r/n2/r/n．求證：/r/n（Ⅰ）數列/r/n{/r/na/r/nn/r/n﹣/r/nb/r/nn/r/n}/r/n是等差數列；/r/n（Ⅱ）/r/n…/r/n+/r/n≤/r/n2/r/n（/r/n1/r/n﹣/r/n）．/r/n證明：（Ⅰ）數列/r/n{/r/na/r/nn/r/n}/r/n是各項均為正數的等比數列，若/r/na/r/n1/r/n＝/r/n2/r/n，/r/na/r/n2/r/n+/r/na/r/n3/r/n是/r/na/r/n3/r/n與/r/na