数列求和（必修5第二章）

单击此处编辑母版标题样式,单击此处编辑母版文本样式,第二级,第三级,第四级,第五级,*,中央电视台的,开心辞典,栏目，有一次,的最后一题是：“给出一组数,1,，,3,，,6,，,10,，,15,，则第,7,个数是什么？”你认为第,7,个数,是,.,那么，这组数之间的规律是,.,28,a,n,=,n(n+1),2,a,2,-a,1,=2,a,3,-a,2,=3,a,4,-a,3,=4,a,n,-a,n-1,=,n,a,n,=1+2+3+,+n,叠加求和,高考二轮复习-数列求和,知识要点归纳:,重点,:,难点,:,等差、等比数列求和公式,非等差、等比数列的求和,学习目标,:,等差、等比数列的前,n,项和公式和其它几种,常见方法,:,倒序相加法、错位相减法、,a,n,法,(,列,项法、拆项法,).,要深刻理解这些求和方法和含义,熟练掌握它,们适用的数列类型以及在求和中应注意的问题,.,2,：,等比数列前,n,项和公式,:,S,n,=,n(a,1,+a,n,),2,=n,a,1,+d,n(n-1),2,a,1,(1-q,n,),1-q,n,(n+1)(2n+1),6,复习：,1,：,等差数列前,n,项和公式：,S,n,=,a,1,-,a,n,q,1-q,=,(q 1),(q=1),na,1,3:,1,2,+2,2,+3,2,+n,2,=,1,3,+2,3,+3,3,+n,3,=,n,(n+1),2,2,求,数列的前,n,项和，通常要掌握以下解法：,1,直接法,2,公式法,3,倒序相加法,4,错位相减法,5,分组转化法,6,裂项相消法,“a,n,”法,倒序相加法：,例1：,求数列,n,c,100,n,的,前,99,项的和,.,S,99,=c,100,1,+,2c,100,2,+98c,100,98,+99 c,100,99,S,99,=99c,100,99,+98c,100,98,+2c,100,2,+c,100,1,2S,99,=100c,100,1,+,100c,100,2,+100c,100,99,=100(c,100,1,+,c,100,2,+c,100,3,+c,100,99,),=100,(2,100,-2),S,99,=50,(2,100,-2),训练,即时,+2,+3 +n,C,n,2,C,n,1,C,n,n,C,n,3,S,n,=,C,n,2,C,n,1,C,n,n,C,n,n-1,n +(n-1)+,2,+,S,n,=,2S,n,=n(+,+),C,n,0,C,n,1,C,n,n,lim,n,+2,+3 +n,C,n,2,C,n,1,C,n,n,C,n,3,n,3,n,A0 B C2 D,不存在,1,2,=(),原式,=,=n2,n,=0,S,n,=n2,n-1,lim,(),n,n,2,3,1,2,错位相减,求数列,的前,n,项和,.,2n-1,2,n,例2,S,n,=+,+,1,2,3,2,2,5,2,3,2n-1,2,n,S,n,=,1,2,1,2,2,3,2,3,5,2,4,2n-1,2,n+1,+,+,(1),(2),S,n,=,1,2,1,2,2,1,2,3,1,2,n-1,+,+,(1)-(2),得：,1,2,1,2,+,+,+,2n-1,2,n+1,-,（）,=,3,2,-,2n+3,2,n+1,S,n,=,3-,2n+3,2,n,解：,练习,:,求数列,(2n+1)2,n-1,的前,n,项和,.,训练,即时,S,n,=3,2,0,+5,2,1,+72,2,+92,3,+(2n+1)2,n-1,2S,n,=3,2,1,+5,2,2,+72,3,+92,4,+(2n+1)2,n,(1),(2),(1)-(2),得：,-,S,n,=3+,2,2,+2,3,+2,4,+2,n,-(2n+1)2,n,(),=3+2,2,(2,n-1,-1)-(2n+1),2,n,=-1+(1-2n),2,n,S,n,=1+(2n-1),2,n,求和：,+,1,12,1,n(n+1),1,34,1,23,例3：,1,n(n+1),a,n,=,解：,=-,1,n,1,n+1,S,n,=+,1,12,1,n(n+1),1,34,1,23,+(-),1,n,1,n+1,=(1-)+(-)+(-)+,1,2,1,2,1,3,1,3,1,4,1,n+1,=1-,=,n,n+1,“an”法,训练,即时,1,求：,1,、,1+2,、,1+2+2,2,、,1+2+2,2,+2,3,的,前,n,项和,.,a,n,=1+2+2,2,+,+2,n-1,=,1-2,1-2,n,=2,n,-1,S,n,=,1+(1+2)+(1+2+2,2,)+(1+2+2,2,+2,3,),+,+,(,1+2+2,2,+,+2,n-1,),=(2-1)+(2,2,-1)+(2,3,-1)+(2,4,-1)+,+(2,n,-1),=(2+2,2,+2,3,+,+2,n,)-n,=2,n+1,-n-2,2,求和,：,+,1,24,1,13,1,35,1,n(n+2),3,求和,:1 +,1,1+2,1,1+2+3,1,1+2+.+n,2n,n+1,2,1,(1+-),2,1,1,n+2,1,n+1,某射手射击目标，直到击中目标才能停止，,若某射手射击到第,n,次击中目标，设该射手,击中目标的概率为,p,，,试求：,(1),射击次数,(n),的概率分布列；,(2),求射击,n,次的数学期望,.,(1),设射击次数,所取值为,1,、,2,、,3,、,、,n.,1 2 3,n,p(,),P,(1-p)p,(1-p),2,p,(1-p),n-1,p,(2)E,=p+2(1-p)p+3(1-p),2,p+,+n(1-p),n-1,p,=p1+2(1-p)+3(1-p),2,+,+n(1-p),n-1,E,=,p,1-(1-p),n,(1+np),解,:,例4：,n,2,个正数排成如下表示的,n,行,n,列：,a,11,a,12,a,13,a,14,a,1n,a,21,a,22,a,23,a,24,a,2n,a,n1,a,n2,a,n3,a,n4,a,nn,其中每一行成等差数列，每一列成等比数,列,且各列公比相等,若,a,24,=1,a,42,=1/8,a,43,=3/16,求,a,11,+a,22,+a,33,+a,44,+,+,a,nn,的值,.,分析：,探究,a,nn,的表达式,.,例5：,解：,设第一行数列公差为,d,各列的公比为,q.,a,24,=(a,11,+3d)q=1,a,42,=(a,11,+d)q,3,=,1,8,解得,:,a,11,=d=q=,1,2,(,负值舍去）,a,nn,=a,1n,q,n-1,=a,11,+(n-1)dq,n-1,n,2,n,=,S,n,=+,+,1,2,2,2,2,3,2,3,n,2,n,错位相减,S,n,=,2-,n+2,2,n,a,43,=+dq,3,=,3,16,1,8,则第四行数列的公差为,dq,3,走向高考赛场:,已知数列,a,n,的前,n,项和为,S,n,且对任意正整数,n,都,有,2S,n,=(n+2)a,n,-1.,求数列,a,n,的通项公式,;,设,且,T,n,m,恒成立,求,m,的最小值,.,m,的最小值是,一般数列求和方法总结,:,1,、直接由等差、等比数列的求和公式求和，注意等比时,q=1,q1,的讨论,.,2,、倒序相加法,;,3,、错位相减法,;,4,、“,a,n,”,法,.,小结：,作业：,综合试卷,