2017 2018高中数学第二章数列23等差数列的前n项和1新人教A必修5

1、2.3,等差数列的前,n,项和,一,1,掌握等差数列前,n,项和公式及其获取思路,2,经历公式的推导过程，体验从特殊到一般的研究方法，学,会观察、归纳、反思,3,熟练掌握等差数列的五个量,a,1,d,n,a,n,S,n,的关系，能够,由其中三个求另外两个,学习目标,题型探究,问题导学,内容索引,当堂训练,问题导学,思考,知识点一,等差数列前,n,项和公式的推导,高斯用,1,2,3,100,1,100,2,99,50,51,101,50,迅速求出了等差数列前,100,项的和但如果是求,1,2,3,n,不知道共有奇数项还是偶数项怎么办,答案,倒序相加法,可以推广到一般等差数列求前,n,项和，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,两式相加，得,2,S,n,n,a,1,a,n,梳理,由此可得等差数列,a,n,的前,n,项和公式,S,n,n,a,1,a,n,2,根据等差数列的通项公式,a,n,a,1,n,1,d,n,n,1,2,d,代入上式可得,S,n,na,1,_,知识点二,等差数列前,n,项和公式的特征,思考,1,答案,等差数列,a,n,中，若已知,a,2,

3、7,能求出前,3,项和,S,3,吗,S,3,3,a,1,a,3,2,3,a,1,a,3,2,3,a,2,21,思考,2,我们对等差数列的通项公式变形,a,n,a,1,n,1,d,dn,a,1,d,分析出通项公式与一次函数的关系你能类比这,个思路分析一下,S,n,na,1,n,n,1,2,d,吗,按,n,的降幂展开,S,n,na,1,n,n,1,2,d,d,2,n,2,a,1,d,2,n,是关于,n,的二次函数形式，且常数项为,0,答案,梳理,等差数列,a,n,的前,n,项和,S,n,有下面几种常见变形,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,是公差为,d,2,的等差数列,知识点三,等差数列前,n,项和公式的性质,思考,如果,a,n,是等差数列，那么,a,1,a,2,a,10,a,11,a,12,a,20,a,21,a,22,a,30,是等差数列吗,答案,梳理,1,S,m,S,2,m,S,3,m,分别为等差数列,a,n,的前,m,项，前,2,m,项，前,3,m,项的和,则,S,m,S,2,m,S,m,S,3,m,S,2,m,也成等差数列，公差为,m,2,d,2,若等差数列的项数为,2,n,n,N,则,S,2,n,且,S,偶,S,奇,3,若等差数列的项数为,2,n,1,n,N,则,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,题型探究,命题角度,1,方程思想,例,1,已知一个等差数列,a,n,的前,10,项的和是,310,前,20,项的和是,1,220,由这些条件能确定这个等差数列的前,n,项和的公式吗,类型一,等差数列前,n,项和公式的应用,解答,1,在解决与等差数列前,n,项和有关的问题时，要注意方程思想和整体,思想的运用,2,构成等差数列前,n,项和公式的元素有,a,1,d,n,a,n,S,n,知其三能,求其二,反思与

6、感悟,跟踪训练,1,在等差数列,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,解方程组得,n,5,a,1,3,或,n,7,a,1,1,命题角度,2,实际应用,例,2,某人用分期付款的方式购买一件家电，价格为,1,150,元，购买当天,先付,150,元，以后每月的这一天都交付,50,元，并加付欠款利息，月利率,为,1,若交付,150,元后的一个月开始算分期付款的第一个月，则分期付,款的第,10,个月该交付多少钱？全部贷款付清后，

7、买这件家电实际花费,多少钱,解答,建立等差数列的模型时，要根据题意找准首项、公差和项数或者首项,末项和项数本题是根据首项和公差选择前,n,项和公式进行求解,反思与感悟,跟踪训练,2,甲、乙两物体分别从相距,70,m,的两处同时相向运动，甲第,1,分钟走,2,m,以后每分钟比前,1,分钟多走,1,m,乙每分钟走,5,m,1,甲、乙开始运动后几分钟相遇,有,2,n,n,n,1,2,5,n,70,整理得,n,2,13,n,140,0,设,n,分钟后第,1,次相遇，依题意,解答,解之得,n,7,n,20,舍去,所以第,1,次相遇是在开始运动后,7,分钟,2,如果甲、乙到达对方起点后立即返回，甲继续每分

8、钟比前,1,分钟多走,1,m,乙继续每分钟走,5,m,那么开始运动几分钟后第二次相遇,有,2,n,n,n,1,2,5,n,3,70,设,n,分钟后第,2,次相遇，依题意,解答,整理得,n,2,13,n,420,0,解之得,n,15,n,28,舍去,所以第,2,次相遇是在开始运动后,15,分钟,类型二,等差数列前,n,项和的性质的应用,例,3,1,等差数列,a,n,的前,m,项和为,30,前,2,m,项和为,100,求数列,a,n,的,前,3,m,项的和,S,3,m,解答,2,两个等差数列,a,n,b,n,的前,n,项和分别为,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,反思与感悟,等差数列前,n,项和,S,n,的有关性质在解题过程中，如果运用得当可以达,到化繁为简、化难为易、事半功倍的效果,跟踪训练,3,设,a,n,为等差数列,S,n,为数列,a,n,的前,n,项和，已知,S,7,7,S,15,75,T,n,为数列,的前,n,项和，求,T,n,S,n,n,解答,当堂训练,1,在等差数列,a,n,中，若,S,10,120,则,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,记等差数列的前,n,项和为,S,n,若,S,2,4,S,4,20,则该数列的公差,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,在一个等差数列中，已知,a,10,10,则,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,已知等差数列,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,规律与方法,1,求等差数列前,n,项和公式的方法称为倒序相加法，在某些数列求和中,也可能用到,2,等差数列的两个求和