數(shù)列全章復(fù)習(xí)-公開課
單擊此處編輯母版標(biāo)題樣式,單擊此處編輯母版文本樣式,第二級(jí),第三級(jí),第四級(jí),第五級(jí),*,數(shù)列 復(fù)習(xí)課,數(shù)列,通項(xiàng)a,n,等差數(shù)列,前n項(xiàng)和S,n,等比數(shù)列,定義,通 項(xiàng),前n項(xiàng)和,性 質(zhì),知識(shí),結(jié)構(gòu),一、知識(shí)回顧,等 差 數(shù) 列,等 比 數(shù) 列,定 義,通 項(xiàng),中 項(xiàng),性 質(zhì),求和公式,關(guān)系式,適用所有數(shù)列,a,A,b成等差數(shù)列,則,a,G,b成等比數(shù)列,則,若m+n=p+q則,若m+n=p+q則,仍成等差,仍成等比,牛刀小試,在等差數(shù)列a,n,中,,,a,2,=-2,a,5,=54,,求,a,8=_.,在等差數(shù)列a,n,中,,,若,a,3,+a,4,+a,5,+a,6,+a,7,=450,,則,a,2,+a,8,的值為_.,在等差數(shù)列a,n,中,,,a,15,=10,a,45,=90,則,a,60,=_.,在等差數(shù)列a,n,中,,a,1,+a,2,=30,a,3,+a,4,=120,則,a,5,+a,6,=_,.,110,運(yùn)用性質(zhì):,a,n,=a,m,+,(n-m)d或等差中項(xiàng),運(yùn)用性質(zhì):,若,n+m=p+q,則,a,m,+a,n,=,a,p,+,a,q,運(yùn)用性質(zhì):,從原數(shù)列中取出偶數(shù)項(xiàng)組成的新數(shù)列公差為2d.(可推廣),運(yùn)用性質(zhì):,若,a,n是公差為,d,的等差數(shù)列,c,n,是公差為d,的,等差數(shù)列,,則數(shù)列,a,n,+,c,n,是公差為,d+d,的等差數(shù)列。,180,130,210,在等比數(shù)列a,n,中,,a,2,=-2,a,5,=54,,a,8=,.,在等比數(shù)列a,n,中,,且a,n,0,,a,2,a,4,+2a,3,a,5,+a,4,a,6,=36,那么a,3,+a,5,=,_,.,在等比數(shù)列a,n,中,,a,15,=10,a,45,=90,則,a,60,=_.,在等比數(shù)列a,n,中,a,1,+a,2,=30,a,3,+a,4,=120,則a,5,+a,6,=_,.,-1458,6,270,480,或-270,牛刀小試,常見的求和公式,專題一:一般數(shù)列求和法,倒序相加法,求和,如a,n,=3n+1,錯(cuò)項(xiàng)相減法,求和,如a,n,=(2n-1)2,n,分組法,求和,如a,n,=2n+3,n,裂項(xiàng)相加法,求和,如a,n,=1/n(n+1),公式法,求和,如a,n,=2n,2,-5n,專題一:一般數(shù)列求和法,一、倒序相加法,解:,例1:,二、錯(cuò)位相減法,解,:,“錯(cuò)位相減法”,求和,常應(yīng)用于形如a,n,b,n,的數(shù)列,求和,其中,a,n,為等,差,數(shù)列,b,n,為等,比,數(shù)列,b,n,的公比為q,則可借助 轉(zhuǎn)化為等比數(shù)列,的求和問題。,三、分組求和,把數(shù)列的,每一項(xiàng)分成幾項(xiàng),,或把數(shù)列的,項(xiàng)“集”在一塊重新組合,,或,把整個(gè)數(shù)列分成幾部分,,使其轉(zhuǎn)化為等差或等比數(shù)列,這一求和方法稱為分組轉(zhuǎn)化法.,練習(xí):求和,解:,四、裂項(xiàng)相消求和法,常用列項(xiàng)技巧:,把數(shù)列的通項(xiàng)拆成兩項(xiàng)之,差,,即數(shù)列的每一項(xiàng)都可按,此法拆成兩項(xiàng)之,差,,在求和時(shí)一些正負(fù)項(xiàng)相互抵消,,于是前n項(xiàng)的和變成首尾若干少數(shù)項(xiàng)之和,這一求和,方法稱為裂項(xiàng)相消法.,累加,法,如,累乘,法,如,構(gòu)造新數(shù)列,:如,取倒數(shù),:如,S,n,和a,n,的關(guān)系,:,專題二:通項(xiàng)的求法,數(shù)列的前,n,項(xiàng)和,S,n,n,2,n,+1,,則通項(xiàng),a,n,=_,-得,:,1、數(shù)列1,7,13,19的一個(gè)通項(xiàng)公式為(),A、,a,n,=2,n,1,B、,a,n,=6,n,+5,C、,a,n,=(1),n,6,n,5,D、,a,n,=,(1),n,(6,n,5),D,2.數(shù)列,a,n,的前n項(xiàng)和,S,n,=n,2,+,1,則,a,n,=,_,.,3、寫出下列數(shù)列的一個(gè)通項(xiàng)公式,(1)、,(2)、,解:(1)、注意分母是 ,分,子比分母少1,故,(2)、由奇數(shù)項(xiàng)特征及偶數(shù)項(xiàng)特征得,返回,4、在各項(xiàng)均為正數(shù)的等比數(shù)列a,n,中,若a,5,a,6,=9,則log,3,a,1,+log,3,a,2,+log,3,a,10,等于(),(A)12(B,),10(C)8(D)2+log,3,5,B,5、等差數(shù)列a,n,的各項(xiàng)都是小于零的數(shù),且 ,則它的前10項(xiàng)和S,10,等于(),(A)-9(B)-11(C)-13(D)-15,D,6、在公比q1的等比數(shù)列a,n,中,若a,1,+a,4,=18,a,2,+a,3,=12,則這個(gè)數(shù)列的前8項(xiàng)之和S,8,等于(),(A)513(B)512(C)510(D),C,7,、在數(shù)列a,n,中,a,n+1,=Ca,n,(C為非零常數(shù))且前n項(xiàng)和S,n,=3,n,+k則k等于(),(A)-1(B)1(C)0(D)2,A,8,、等差數(shù)列a,n,中,若S,m,=S,n,(m,n),則S,m+n,的值為(),D,9,、等差數(shù)列a,n,是遞減數(shù)列,a,2,a,3,a,4,=48,a,2,+a,3,+a,4,=12,則數(shù)列a,n,的通項(xiàng)公式(),(A)a,n,=2n-2(B)a,n,=2n+2,(C)a,n,=-2n+12(D)a,n,=-2n+10,D,1,0,、在等差數(shù)列a,n,中,a,1,+3a,8,+a,15,=120,則2a,9,-a,10,的值為(),(A)24(B)22(C)2(D)-8,A,考點(diǎn)練習(xí),1、在等比數(shù)列,a,n,中,,a,3,a,4,a,5,=3,,a,6,a,7,a,8,=24,則,a,9,a,10,a,11,的值等于_,192,考點(diǎn)練習(xí),2、,a,=,,b,=,,a,、,b,的,等差中項(xiàng)為(),A、B、,C、D、,A,3、設(shè),a,n,為等差數(shù)列,,S,n,為前,n,項(xiàng),和,,a,4,=,,S,8,=4,求,a,n,與,S,n,點(diǎn)評(píng):,在等差數(shù)列中,由a,1,、d、n、a,n,、s,n,知三求二,考點(diǎn)練習(xí),4、數(shù)列,a,n,滿足,a,1,=,,a,1,+,a,2,+,a,3,+,a,n,=,n,2,a,n,,求通項(xiàng),a,n,解析:a,1,+a,2,+a,3,+a,n,=n,2,a,n,a,1,+a,2,+a,n-1,=(n-1),2,a,n-1,(n2),相減 a,n,=n,2,a,n,-(n-1),2,a,n-1,考點(diǎn)練習(xí),