2017 2018高中數(shù)學(xué)第二章數(shù)列23等差數(shù)列的前n項(xiàng)和1新人教A必修5

1、2.3,等差數(shù)列的前,n,項(xiàng)和,一,1,掌握等差數(shù)列前,n,項(xiàng)和公式及其獲取思路,2,經(jīng)歷公式的推導(dǎo)過程，體驗(yàn)從特殊到一般的研究方法，學(xué),會(huì)觀察、歸納、反思,3,熟練掌握等差數(shù)列的五個(gè)量,a,1,d,n,a,n,S,n,的關(guān)系，能夠,由其中三個(gè)求另外兩個(gè),學(xué)習(xí)目標(biāo),題型探究,問題導(dǎo)學(xué),內(nèi)容索引,當(dāng)堂訓(xùn)練,問題導(dǎo)學(xué),思考,知識(shí)點(diǎn)一,等差數(shù)列前,n,項(xiàng)和公式的推導(dǎo),高斯用,1,2,3,100,1,100,2,99,50,51,101,50,迅速求出了等差數(shù)列前,100,項(xiàng)的和但如果是求,1,2,3,n,不知道共有奇數(shù)項(xiàng)還是偶數(shù)項(xiàng)怎么辦,答案,倒序相加法,可以推廣到一般等差數(shù)列求前,n,項(xiàng)和，其方法

2、如下,S,n,a,1,a,2,a,3,a,n,1,a,n,a,1,a,1,d,a,1,2,d,a,1,n,2,d,a,1,n,1,d,S,n,a,n,a,n,1,a,n,2,a,2,a,1,a,n,a,n,d,a,n,2,d,a,n,n,2,d,a,n,n,1,d,兩式相加，得,2,S,n,n,a,1,a,n,梳理,由此可得等差數(shù)列,a,n,的前,n,項(xiàng)和公式,S,n,n,a,1,a,n,2,根據(jù)等差數(shù)列的通項(xiàng)公式,a,n,a,1,n,1,d,n,n,1,2,d,代入上式可得,S,n,na,1,_,知識(shí)點(diǎn)二,等差數(shù)列前,n,項(xiàng)和公式的特征,思考,1,答案,等差數(shù)列,a,n,中，若已知,a,2,

3、7,能求出前,3,項(xiàng)和,S,3,嗎,S,3,3,a,1,a,3,2,3,a,1,a,3,2,3,a,2,21,思考,2,我們對(duì)等差數(shù)列的通項(xiàng)公式變形,a,n,a,1,n,1,d,dn,a,1,d,分析出通項(xiàng)公式與一次函數(shù)的關(guān)系你能類比這,個(gè)思路分析一下,S,n,na,1,n,n,1,2,d,嗎,按,n,的降冪展開,S,n,na,1,n,n,1,2,d,d,2,n,2,a,1,d,2,n,是關(guān)于,n,的二次函數(shù)形式，且常數(shù)項(xiàng)為,0,答案,梳理,等差數(shù)列,a,n,的前,n,項(xiàng)和,S,n,有下面幾種常見變形,1,S,n,n,a,1,a,n,2,2,S,n,d,2,n,2,a,1,d,2,n,3,S,

4、n,n,d,2,n,a,1,d,2,S,n,n,是公差為,d,2,的等差數(shù)列,知識(shí)點(diǎn)三,等差數(shù)列前,n,項(xiàng)和公式的性質(zhì),思考,如果,a,n,是等差數(shù)列，那么,a,1,a,2,a,10,a,11,a,12,a,20,a,21,a,22,a,30,是等差數(shù)列嗎,答案,梳理,1,S,m,S,2,m,S,3,m,分別為等差數(shù)列,a,n,的前,m,項(xiàng)，前,2,m,項(xiàng)，前,3,m,項(xiàng)的和,則,S,m,S,2,m,S,m,S,3,m,S,2,m,也成等差數(shù)列，公差為,m,2,d,2,若等差數(shù)列的項(xiàng)數(shù)為,2,n,n,N,則,S,2,n,且,S,偶,S,奇,3,若等差數(shù)列的項(xiàng)數(shù)為,2,n,1,n,N,則,S,2

5、,n,1,且,S,奇,S,偶,a,n,S,奇,na,n,S,偶,n,1,a,n,S,奇,S,偶,a,n,a,n,1,S,奇,S,偶,n,n,1,n,a,n,a,n,1,nd,2,n,1,a,n,題型探究,命題角度,1,方程思想,例,1,已知一個(gè)等差數(shù)列,a,n,的前,10,項(xiàng)的和是,310,前,20,項(xiàng)的和是,1,220,由這些條件能確定這個(gè)等差數(shù)列的前,n,項(xiàng)和的公式嗎,類型一,等差數(shù)列前,n,項(xiàng)和公式的應(yīng)用,解答,1,在解決與等差數(shù)列前,n,項(xiàng)和有關(guān)的問題時(shí)，要注意方程思想和整體,思想的運(yùn)用,2,構(gòu)成等差數(shù)列前,n,項(xiàng)和公式的元素有,a,1,d,n,a,n,S,n,知其三能,求其二,反思與

6、感悟,跟蹤訓(xùn)練,1,在等差數(shù)列,a,n,中，已知,d,2,a,n,11,S,n,35,求,a,1,和,n,由,a,n,a,1,n,1,d,S,n,na,1,n,n,1,2,d,解答,得,a,1,2,n,1,11,na,1,n,n,1,2,2,35,解方程組得,n,5,a,1,3,或,n,7,a,1,1,命題角度,2,實(shí)際應(yīng)用,例,2,某人用分期付款的方式購(gòu)買一件家電，價(jià)格為,1,150,元，購(gòu)買當(dāng)天,先付,150,元，以后每月的這一天都交付,50,元，并加付欠款利息，月利率,為,1,若交付,150,元后的一個(gè)月開始算分期付款的第一個(gè)月，則分期付,款的第,10,個(gè)月該交付多少錢？全部貸款付清后，

7、買這件家電實(shí)際花費(fèi),多少錢,解答,建立等差數(shù)列的模型時(shí)，要根據(jù)題意找準(zhǔn)首項(xiàng)、公差和項(xiàng)數(shù)或者首項(xiàng),末項(xiàng)和項(xiàng)數(shù)本題是根據(jù)首項(xiàng)和公差選擇前,n,項(xiàng)和公式進(jìn)行求解,反思與感悟,跟蹤訓(xùn)練,2,甲、乙兩物體分別從相距,70,m,的兩處同時(shí)相向運(yùn)動(dòng)，甲第,1,分鐘走,2,m,以后每分鐘比前,1,分鐘多走,1,m,乙每分鐘走,5,m,1,甲、乙開始運(yùn)動(dòng)后幾分鐘相遇,有,2,n,n,n,1,2,5,n,70,整理得,n,2,13,n,140,0,設(shè),n,分鐘后第,1,次相遇，依題意,解答,解之得,n,7,n,20,舍去,所以第,1,次相遇是在開始運(yùn)動(dòng)后,7,分鐘,2,如果甲、乙到達(dá)對(duì)方起點(diǎn)后立即返回，甲繼續(xù)每分

8、鐘比前,1,分鐘多走,1,m,乙繼續(xù)每分鐘走,5,m,那么開始運(yùn)動(dòng)幾分鐘后第二次相遇,有,2,n,n,n,1,2,5,n,3,70,設(shè),n,分鐘后第,2,次相遇，依題意,解答,整理得,n,2,13,n,420,0,解之得,n,15,n,28,舍去,所以第,2,次相遇是在開始運(yùn)動(dòng)后,15,分鐘,類型二,等差數(shù)列前,n,項(xiàng)和的性質(zhì)的應(yīng)用,例,3,1,等差數(shù)列,a,n,的前,m,項(xiàng)和為,30,前,2,m,項(xiàng)和為,100,求數(shù)列,a,n,的,前,3,m,項(xiàng)的和,S,3,m,解答,2,兩個(gè)等差數(shù)列,a,n,b,n,的前,n,項(xiàng)和分別為,S,n,和,T,n,已知,S,n,T,n,7,n,2,n,3,求,a

9、,5,b,5,的值,a,5,b,5,1,2,a,1,a,9,1,2,b,1,b,9,9,a,1,a,9,2,9,b,1,b,9,2,S,9,T,9,7,9,2,9,3,65,12,解答,反思與感悟,等差數(shù)列前,n,項(xiàng)和,S,n,的有關(guān)性質(zhì)在解題過程中，如果運(yùn)用得當(dāng)可以達(dá),到化繁為簡(jiǎn)、化難為易、事半功倍的效果,跟蹤訓(xùn)練,3,設(shè),a,n,為等差數(shù)列,S,n,為數(shù)列,a,n,的前,n,項(xiàng)和，已知,S,7,7,S,15,75,T,n,為數(shù)列,的前,n,項(xiàng)和，求,T,n,S,n,n,解答,當(dāng)堂訓(xùn)練,1,在等差數(shù)列,a,n,中，若,S,10,120,則,a,1,a,10,的值是,A.12,B.24,C.3

10、6,D.48,答案,解析,由,S,10,10,a,1,a,10,2,1,2,3,4,得,a,1,a,10,S,10,5,120,5,24,2,記等差數(shù)列的前,n,項(xiàng)和為,S,n,若,S,2,4,S,4,20,則該數(shù)列的公差,d,等于,A.2,B.3,C.6,D.7,答案,解析,方法一,由,S,2,2,a,1,d,4,S,4,4,a,1,6,d,20,解得,d,3,方法二,由,S,4,S,2,a,3,a,4,a,1,2,d,a,2,2,d,S,2,4,d,所以,20,4,4,4,d,解得,d,3,1,2,3,4,3,在一個(gè)等差數(shù)列中，已知,a,10,10,則,S,19,_,S,19,19,a,1

11、,a,19,2,19,a,10,a,10,2,19,a,10,19,10,190,1,2,3,4,190,答案,解析,4,已知等差數(shù)列,a,n,中,S,n,n,3,2,1,2,n,n,1,2,15,整理得,n,2,7,n,60,0,解得,n,12,或,n,5,舍去,1,2,3,4,1,a,1,3,2,d,1,2,S,n,15,求,n,及,a,n,解答,a,12,3,2,12,1,1,2,4,n,12,a,n,a,12,4,2,a,1,1,a,n,512,S,n,1,022,求,d,由,S,n,n,a,1,a,n,2,n,1,512,2,1 022,解得,n,4,又由,a,n,a,1,n,1,d,即,512,1,4,1,d,解得,d,171,解答,1,2,3,4,規(guī)律與方法,1,求等差數(shù)列前,n,項(xiàng)和公式的方法稱為倒序相加法，在某些數(shù)列求和中,也可能用到,2,等差數(shù)列的兩個(gè)求和