新高考數(shù)學(xué)一輪復(fù)習(xí)考點精講練+易錯題型第35講 數(shù)列的求和(解析版)_第1頁
新高考數(shù)學(xué)一輪復(fù)習(xí)考點精講練+易錯題型第35講 數(shù)列的求和(解析版)_第2頁
新高考數(shù)學(xué)一輪復(fù)習(xí)考點精講練+易錯題型第35講 數(shù)列的求和(解析版)_第3頁
新高考數(shù)學(xué)一輪復(fù)習(xí)考點精講練+易錯題型第35講 數(shù)列的求和(解析版)_第4頁
新高考數(shù)學(xué)一輪復(fù)習(xí)考點精講練+易錯題型第35講 數(shù)列的求和(解析版)_第5頁
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第35講數(shù)列求和【基礎(chǔ)知識全通關(guān)】1.等差數(shù)列的前n項和首項為SKIPIF1<0,末項為SKIPIF1<0,項數(shù)為n的等差數(shù)列SKIPIF1<0的前n項和公式:SKIPIF1<0.令SKIPIF1<0,SKIPIF1<0,可得SKIPIF1<0,則SKIPIF1<0當(dāng)SKIPIF1<0,即SKIPIF1<0時,SKIPIF1<0是關(guān)于n的二次函數(shù),點SKIPIF1<0是SKIPIF1<0的圖象上一系列孤立的點;SKIPIF1<0當(dāng)SKIPIF1<0,即SKIPIF1<0時,SKIPIF1<0是關(guān)于n的一次函數(shù)SKIPIF1<0,即SKIPIF1<0或常函數(shù)SKIPIF1<0,即SKIPIF1<0,點SKIPIF1<0是直線SKIPIF1<0上一系列孤立的點.我們可以借助二次函數(shù)的圖象和性質(zhì)來研究等差數(shù)列的前n項和的相關(guān)問題.2.用前n項和公式法判定等差數(shù)列等差數(shù)列的前n項和公式與函數(shù)的關(guān)系給出了一種判斷數(shù)列是否為等差數(shù)列的方法:若數(shù)列SKIPIF1<0的前n項和SKIPIF1<0,那么當(dāng)且僅當(dāng)SKIPIF1<0時,數(shù)列SKIPIF1<0是以SKIPIF1<0為首項,SKIPIF1<0為公差的等差數(shù)列;當(dāng)SKIPIF1<0時,數(shù)列SKIPIF1<0不是等差數(shù)列.3.等差數(shù)列的常用性質(zhì)由等差數(shù)列的定義可得公差為SKIPIF1<0的等差數(shù)列SKIPIF1<0具有如下性質(zhì):(1)通項公式的推廣:SKIPIF1<0,SKIPIF1<0.(2)若SKIPIF1<0,則SKIPIF1<0SKIPIF1<0.特別地,①若SKIPIF1<0,則SKIPIF1<0SKIPIF1<0;②若SKIPIF1<0,則SKIPIF1<0SKIPIF1<0.③有窮等差數(shù)列中,與首末兩項等距離的兩項之和都相等,都等于首末兩項的和,即SKIPIF1<0(3)下標(biāo)成等差數(shù)列的項SKIPIF1<0組成以md為公差的等差數(shù)列.(4)數(shù)列SKIPIF1<0是常數(shù)SKIPIF1<0是公差為td的等差數(shù)列.(5)若數(shù)列SKIPIF1<0為等差數(shù)列,則數(shù)列SKIPIF1<0SKIPIF1<0是常數(shù)SKIPIF1<0仍為等差數(shù)列.(6)若SKIPIF1<0,則SKIPIF1<0.4.與等差數(shù)列各項的和有關(guān)的性質(zhì)利用等差數(shù)列的通項公式及前n項和公式易得等差數(shù)列的前n項和具有如下性質(zhì):設(shè)等差數(shù)列SKIPIF1<0(公差為d)和SKIPIF1<0的前n項和分別為SKIPIF1<0,(1)數(shù)列SKIPIF1<0是等差數(shù)列,首項為SKIPIF1<0,公差為SKIPIF1<0.(2)SKIPIF1<0構(gòu)成公差為SKIPIF1<0的等差數(shù)列.(3)若數(shù)列SKIPIF1<0共有SKIPIF1<0項,則SKIPIF1<0,SKIPIF1<0.(4)若數(shù)列SKIPIF1<0共有SKIPIF1<0項,則SKIPIF1<0SKIPIF1<0,SKIPIF1<0SKIPIF1<0.(5)SKIPIF1<0,SKIPIF1<0.5.等比數(shù)列的前n項和公式首項為SKIPIF1<0,公比為SKIPIF1<0的等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和的公式為SKIPIF1<0(1)當(dāng)公比SKIPIF1<0時,因為SKIPIF1<0,所以SKIPIF1<0是關(guān)于n的正比例函數(shù),則數(shù)列SKIPIF1<0的圖象是正比例函數(shù)SKIPIF1<0圖象上的一群孤立的點.(2)當(dāng)公比SKIPIF1<0時,等比數(shù)列的前SKIPIF1<0項和公式是SKIPIF1<0,即SKIPIF1<0SKIPIF1<0,設(shè)SKIPIF1<0,則上式可寫成SKIPIF1<0的形式,則數(shù)列SKIPIF1<0的圖象是函數(shù)SKIPIF1<0圖象上的一群孤立的點.由此可見,非常數(shù)列的等比數(shù)列的前n項和SKIPIF1<0是一個關(guān)于n的指數(shù)型函數(shù)與一個常數(shù)的和,且指數(shù)型函數(shù)的系數(shù)與常數(shù)項互為相反數(shù).6、等比數(shù)列及其前n項和的性質(zhì)若數(shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列,前n項和為SKIPIF1<0,則有如下性質(zhì):(1)若SKIPIF1<0,則SKIPIF1<0;若SKIPIF1<0,則SKIPIF1<0.推廣:SKIPIF1<0SKIPIF1<0若SKIPIF1<0,則SKIPIF1<0.(2)若SKIPIF1<0成等差數(shù)列,則SKIPIF1<0成等比數(shù)列.(3)數(shù)列SKIPIF1<0仍是公比為SKIPIF1<0的等比數(shù)列;數(shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列;數(shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列;若數(shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列,則數(shù)列SKIPIF1<0是公比為SKIPIF1<0的等比數(shù)列.(4)SKIPIF1<0成等比數(shù)列,公比為SKIPIF1<0.(5)連續(xù)相鄰SKIPIF1<0項的和(或積)構(gòu)成公比為SKIPIF1<0或SKIPIF1<0的等比數(shù)列.(6)當(dāng)SKIPIF1<0時,SKIPIF1<0;當(dāng)SKIPIF1<0時,SKIPIF1<0.(7)SKIPIF1<0.(8)若項數(shù)為SKIPIF1<0,則SKIPIF1<0,若項數(shù)為SKIPIF1<0,則SKIPIF1<0.(9)當(dāng)SKIPIF1<0時,連續(xù)SKIPIF1<0項的和(如SKIPIF1<0)仍組成等比數(shù)列(公比為SKIPIF1<0,SKIPIF1<0).注意:這里連續(xù)m項的和均非零.【考點研習(xí)一點通】考點一求解等差數(shù)列的通項及前n項和1.已知數(shù)列SKIPIF1<0中,SKIPIF1<0,當(dāng)SKIPIF1<0時,SKIPIF1<0,求數(shù)列SKIPIF1<0的通項公式.【解析】當(dāng)SKIPIF1<0時,SKIPIF1<0,即SKIPIF1<0,兩邊同時取倒數(shù),得SKIPIF1<0,即SKIPIF1<0,所以數(shù)列SKIPIF1<0是以SKIPIF1<0為首項,SKIPIF1<0為公差的等差數(shù)列,所以SKIPIF1<0,故SKIPIF1<0.【變式1-1】已知SKIPIF1<0為等差數(shù)列SKIPIF1<0的前n項和,且SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前n項和SKIPIF1<0.【解析】(1)設(shè)等差數(shù)列SKIPIF1<0的公差為d,依題意得SKIPIF1<0,解得SKIPIF1<0,則SKIPIF1<0.故數(shù)列SKIPIF1<0的通項公式為SKIPIF1<0.(2)由(1)得SKIPIF1<0,SKIPIF1<0,故數(shù)列SKIPIF1<0的前n項和SKIPIF1<0.考點二數(shù)列SKIPIF1<0的前n項和的求解2.已知數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0.(1)請問數(shù)列SKIPIF1<0是否為等差數(shù)列?如果是,請證明;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項和.【解析】(1)由SKIPIF1<0可得SKIPIF1<0,兩式相減可得SKIPIF1<0于是由SKIPIF1<0可知數(shù)列SKIPIF1<0為等差數(shù)列.(2)記數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,SKIPIF1<0SKIPIF1<0.故數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0.【變式2-1】設(shè)數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式;(2)求數(shù)列SKIPIF1<0的前n項和SKIPIF1<0.【解析】(1)設(shè),且數(shù)列的前項和為,則有.當(dāng)時,;當(dāng)時,.從而,即,解得.(2)設(shè)數(shù)列的前項和為,當(dāng)時,,所以有當(dāng)時,;當(dāng)時,.綜上,SKIPIF1<0.考點三求解等比數(shù)列的通項及前n項和3.若等比數(shù)列SKIPIF1<0的前SKIPIF1<0項和為SKIPIF1<0,且SKIPIF1<05,則SKIPIF1<0等于A.5 B.16C.17 D.25【答案】C【解析】當(dāng)公比SKIPIF1<0時,SKIPIF1<0故公比不為1,當(dāng)公比SKIPIF1<0時,SKIPIF1<0∴SKIPIF1<0,∴SKIPIF1<0,故選C.【名師點睛】本題重點考查了等比數(shù)列的前n項和,注意對公比SKIPIF1<0的分類討論,這是一個易錯點,同時注意首項與公比均不為零.解決本題時,對公比SKIPIF1<0進(jìn)行分類討論,利用前n項和公式及條件,求出SKIPIF1<0,從而得到結(jié)果.【變式3-1】已知等比數(shù)列SKIPIF1<0的各項均為正數(shù),且SKIPIF1<0,SKIPIF1<0.(1)求數(shù)列SKIPIF1<0的通項公式;(2)若數(shù)列SKIPIF1<0滿足:SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0.【解析】(1)設(shè)等比數(shù)列{an}的公比為q,∵a2=6,a3+a4=72,∴6q+6q2=72,即q2+q-12=0,解得q=3或q=-4.又∵an>0,∴q>0,∴q=3,SKIPIF1<0.∴SKIPIF1<0.(2)∵SKIPIF1<0,∴SKIPIF1<0.考點四等比數(shù)列的性質(zhì)的應(yīng)用4.在等比數(shù)列SKIPIF1<0中,SKIPIF1<0是方程SKIPIF1<0的根,則SKIPIF1<0A.SKIPIF1<0B.2C.1D.SKIPIF1<0【答案】A【解析】由等比數(shù)列的性質(zhì)知SKIPIF1<0,故SKIPIF1<0,故選A.【變式4-1】已知等比數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0_______.【答案】140【解析】方法1:由SKIPIF1<0,SKIPIF1<0,易得公比SKIPIF1<0,根據(jù)等比數(shù)列前n項和的性質(zhì),可得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0.方法2:根據(jù)等比數(shù)列前n項和的性質(zhì),可得SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0,所以SKIPIF1<0.方法3:根據(jù)等比數(shù)列前n項和的性質(zhì),可知SKIPIF1<0,SKIPIF1<0,SKIPIF1<0成等比數(shù)列,則SKIPIF1<0,即SKIPIF1<0,解得SKIPIF1<0.【考點易錯】1.等差數(shù)列SKIPIF1<0的前30項之和為50,前50項之和為30,求SKIPIF1<0。【思路】根據(jù)等差數(shù)列前n項公式SKIPIF1<0,整體代入,或者應(yīng)用公式SKIPIF1<0?!窘馕觥糠ㄒ?∵SKIPIF1<0為等差數(shù)列,∴SKIPIF1<0,∴SKIPIF1<0(2)-(1)有SKIPIF1<0,即SKIPIF1<0∴SKIPIF1<0。法二:∵SKIPIF1<0為等差數(shù)列,∴SKIPIF1<0,∴SKIPIF1<0即SKIPIF1<0∴(2)-(1)有:SKIPIF1<0即SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0。法三:∵SKIPIF1<0為等差數(shù)列,∴SKIPIF1<0,SKIPIF1<0,∵SKIPIF1<0,SKIPIF1<0,…,SKIPIF1<0也為等差數(shù)列,∴SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0.【總結(jié)】法一、二均可用方程思想求出A、B、SKIPIF1<0、d來,然后求未知,運算量則相對很大,此時要注意整體思想的運用。2.設(shè)SKIPIF1<0為數(shù)列SKIPIF1<0的前n項和,且SKIPIF1<0.求證:數(shù)列SKIPIF1<0為等差數(shù)列.【思路】判斷一個數(shù)列是否等差數(shù)列,可以參考考點梳理中羅列的方法。證明:由SKIPIF1<0得SKIPIF1<0,所以SKIPIF1<0整理得SKIPIF1<0,又得SKIPIF1<0相減并整理得:SKIPIF1<0所以數(shù)列SKIPIF1<0是個等差數(shù)列3.設(shè){an}是等差數(shù)列,證明以bn=SKIPIF1<0(n∈N*)為通項公式的數(shù)列{bn}是等差數(shù)列.證法一:設(shè)等差數(shù)列{an}的公差是d(常數(shù)),當(dāng)n≥2時,SKIPIF1<0=SKIPIF1<0-SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0=SKIPIF1<0(常數(shù))∴{bn}是等差數(shù)列.證法二:等差數(shù)列{an}的前n項和SKIPIF1<0,∴bn=SKIPIF1<0∴{bn}是等差數(shù)列.【總結(jié)】判斷或證明數(shù)列是等差數(shù)列的方法有:(1)定義法:an+1-an=d(常數(shù))(n∈N*)SKIPIF1<0{an}是等差數(shù)列;(2)中項公式法:2an+1=an+an+2(n∈N*)SKIPIF1<0{an}是等差數(shù)列;(3)通項公式法:an=kn+b(k、b是常數(shù))(n∈N*)SKIPIF1<0{an}是等差數(shù)列;(4)前n項和公式法:Sn=An2+Bn(A、B是常數(shù))(n∈N*)SKIPIF1<0{an}是等差數(shù)列.4.等差數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0,若SKIPIF1<0,SKIPIF1<0,SKIPIF1<0.(1)求公差d的取值范圍;(2)n為何值時,Sn最大,并說明理由?!窘馕觥浚?)由SKIPIF1<0又由SKIPIF1<0得SKIPIF1<0代入不等式組∴SKIPIF1<0,解出SKIPIF1<0(2)方法一:由(1)知:SKIPIF1<0且SKIPIF1<0∴數(shù)列SKIPIF1<0是遞減數(shù)列,由SKIPIF1<0得SKIPIF1<0∴SKIPIF1<0即SKIPIF1<0,∴SKIPIF1<0中最后一個正數(shù)項是SKIPIF1<0,SKIPIF1<0開始為負(fù)數(shù)項∴當(dāng)n=6時,SKIPIF1<0最大.方法二:由(1)知:SKIPIF1<0且SKIPIF1<0∴數(shù)列SKIPIF1<0是遞減數(shù)列,若要SKIPIF1<0最大,需確定數(shù)列中最后一個非負(fù)數(shù)項是第幾項.由SKIPIF1<0∴SKIPIF1<0即SKIPIF1<0,∴SKIPIF1<0由SKIPIF1<0,∴SKIPIF1<0,即SKIPIF1<0,∴SKIPIF1<0,∴SKIPIF1<0∴SKIPIF1<0中最后一個正數(shù)項是SKIPIF1<0,SKIPIF1<0開始為負(fù)數(shù)項∴當(dāng)n=6時,SKIPIF1<0最大.方法三:SKIPIF1<0SKIPIF1<0∵d<0,∴當(dāng)SKIPIF1<0最小時SKIPIF1<0有最大值,當(dāng)SKIPIF1<0時,SKIPIF1<0∴當(dāng)n=6時SKIPIF1<0最小,即SKIPIF1<0最大,方法四:SKIPIF1<0是等差數(shù)列,故設(shè)SKIPIF1<0,如圖所示∵SKIPIF1<0,SKIPIF1<0,∴拋物線與x軸的另一個交點在n=12與n=13之間?!鄬ΨQ軸l的位置在6與6.5之間,易知n=6對應(yīng)的A點與對稱軸的距離比n=7對應(yīng)的點B與對稱軸的距離要近,故A為最高點,SKIPIF1<0最大。5.若數(shù)列SKIPIF1<0滿足SKIPIF1<0,則稱數(shù)列SKIPIF1<0為“平方遞推數(shù)列”.已知數(shù)列SKIPIF1<0中,SKIPIF1<0,點SKIPIF1<0在函數(shù)SKIPIF1<0的圖象上,其中n為正整數(shù).(1)證明:數(shù)列SKIPIF1<0是“平方遞推數(shù)列”,且數(shù)列SKIPIF1<0為等比數(shù)列;(2)設(shè)(1)中“平方遞推數(shù)列”的前n項之積為SKIPIF1<0,求SKIPIF1<0;(3)在(2)的條件下,記SKIPIF1<0,設(shè)數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0,求使SKIPIF1<0成立的n的最小值.【答案】(1)見解析;(2)SKIPIF1<0;(3)SKIPIF1<0.【解析】(1)由題意得SKIPIF1<0,即SKIPIF1<0,則是“平方遞推數(shù)列”.對SKIPIF1<0兩邊取對數(shù)得SKIPIF1<0,所以數(shù)列SKIPIF1<0是以SKIPIF1<0為首項,2為公比的等比數(shù)列.(2)由(1)知SKIPIF1<0,則SKIPIF1<0(3)由(2)知SKIPIF1<0,SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0,又SKIPIF1<0,所以SKIPIF1<0,故使SKIPIF1<0成立的n的最小值為SKIPIF1<0.6.若數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0滿足SKIPIF1<0.(1)求證:數(shù)列SKIPIF1<0是等比數(shù)列;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項和SKIPIF1<0.【答案】(1)見解析;(2)SKIPIF1<0.【解析】(1)當(dāng)SKIPIF1<0時,SKIPIF1<0,計算得出SKIPIF1<0,當(dāng)SKIPIF1<0時,根據(jù)題意得,SKIPIF1<0,所以SKIPIF1<0,即SKIPIF1<0.SKIPIF1<0,即SKIPIF1<0,SKIPIF1<0數(shù)列SKIPIF1<0是首項為?2,公比為2的等比數(shù)列.(2)由(1)知,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0.【名師點睛】本題考查了等比數(shù)列的證明,數(shù)列求和的常用方法;數(shù)列求和的常用方有:分組求和,用于當(dāng)數(shù)列中相鄰兩項的和或者差是定值的;錯位相減法,用于一個等比數(shù)列和等差數(shù)列乘到一起;裂項相消法主要用于分式型的通項.7.已知等比數(shù)列SKIPIF1<0滿足SKIPIF1<0.(1)求SKIPIF1<0的通項公式;(2)設(shè)SKIPIF1<0,求數(shù)列SKIPIF1<0的前SKIPIF1<0項和.【答案】(1)SKIPIF1<0;(2)SKIPIF1<0.【解析】(1)設(shè)SKIPIF1<0,依題意,有SKIPIF1<0,解得SKIPIF1<0.所以SKIPIF1<0.(2)SKIPIF1<0SKIPIF1<0.記數(shù)列SKIPIF1<0的前n項的和為SKIPIF1<0,則SKIPIF1<0,SKIPIF1<0.兩式相減,得SKIPIF1<0.故SKIPIF1<0.【名師點睛】本題主要考查了數(shù)列通項的求法以及數(shù)列前SKIPIF1<0項和的求法.數(shù)列通項的求法常用的方法有:公式法、累加、累乘等.求數(shù)列前SKIPIF1<0項和的常用的方法有:錯位相減、裂項相消、分組求和等.(1)把SKIPIF1<0和SKIPIF1<0換成SKIPIF1<0和SKIPIF1<0的關(guān)系即可.(2)首先利用裂項把SKIPIF1<0計算出來,再根據(jù)錯位相減即可得出SKIPIF1<0的前SKIPIF1<0項和.【鞏固提升】1.SKIPIF1<0和SKIPIF1<0是兩個等差數(shù)列,其中SKIPIF1<0為常值,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,則SKIPIF1<0()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】B【分析】由已知條件求出SKIPIF1<0的值,利用等差中項的性質(zhì)可求得SKIPIF1<0的值.【詳解】由已知條件可得SKIPIF1<0,則SKIPIF1<0,因此,SKIPIF1<0.故選:B.2.已知數(shù)列SKIPIF1<0滿足SKIPIF1<0.記數(shù)列SKIPIF1<0的前n項和為SKIPIF1<0,則()A.SKIPIF1<0 B.SKIPIF1<0 C.SKIPIF1<0 D.SKIPIF1<0【答案】A【分析】顯然可知,SKIPIF1<0,利用倒數(shù)法得到SKIPIF1<0,再放縮可得SKIPIF1<0,由累加法可得SKIPIF1<0,進(jìn)而由SKIPIF1<0局部放縮可得SKIPIF1<0,然后利用累乘法求得SKIPIF1<0,最后根據(jù)裂項相消法即可得到SKIPIF1<0,從而得解.【詳解】因為SKIPIF1<0,所以SKIPIF1<0,SKIPIF1<0.由SKIPIF1<0SKIPIF1<0,即SKIPIF1<0根據(jù)累加法可得,SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時取等號,SKIPIF1<0SKIPIF1<0,由累乘法可得SKIPIF1<0,當(dāng)且僅當(dāng)SKIPIF1<0時取等號,由裂項求和法得:所以SKIPIF1<0,即SKIPIF1<0.故選:A.【點睛】本題解題關(guān)鍵是通過倒數(shù)法先找到SKIPIF1<0的不等關(guān)系,再由累加法可求得SKIPIF1<0,由題目條件可知要證SKIPIF1<0小于某數(shù),從而通過局部放縮得到SKIPIF1<0的不等關(guān)系,改變不等式的方向得到SKIPIF1<0,最后由裂項相消法求得SKIPIF1<0.3.某校學(xué)生在研究民間剪紙藝術(shù)時,發(fā)現(xiàn)剪紙時經(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.【答案】5SKIPIF1<0【分析】(1)按對折列舉即可;(2)根據(jù)規(guī)律可得SKIPIF1<0,再根據(jù)錯位相減法得結(jié)果.【詳解】(1)由對折2次共可以得到SKIPIF1<0,SKIPIF1<0,SKIPIF1<0三種規(guī)格的圖形,所以對著三次的結(jié)果有:SKIPIF1<0,共4種不同規(guī)格(單位SKIPIF1<0;故對折4次可得到如下規(guī)格:SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,SKIPIF1<0,共5種不同規(guī)格;(2)由于每次對著后的圖形的面積都減小為原來的一半,故各次對著后的圖形,不論規(guī)格如何,其面積成公比為SKIPIF1<0的等比數(shù)列,首項為120SKIPIF1<0,第n次對折后的圖形面積為SKIPIF1<0,對于第n此對折后的圖形的規(guī)格形狀種數(shù),根據(jù)(1)的過程和結(jié)論,猜想為SKIPIF1<0種(證明從略),故得猜想SKIPIF1<0,設(shè)SKIPIF1<0,則SKIPIF1<0,兩式作差得:SKIPIF1<0SKIPIF1<0SKIPIF1<0,因此,SKIPIF1<0.故答案為:SKIPIF1<0;SKIPIF1<0.【點睛】方法點睛:數(shù)列求和的常用方法:(1)對于等差等比數(shù)列,利用公式法可直接求解;(2)對于SKIPIF1<0結(jié)構(gòu),其中SKIPIF1<0是等差數(shù)列,SKIPIF1<0是

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