人教版高中数学等差数列课件

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,*,人教版高中数学等差数列课件,一、概念与公式,一、概念与公式,二、等差数列的性质,1.,首尾项性质,:,有穷等差数列中,与首末两项距离相等的两项和相等,即,:,特别地,若项数为奇数,还等于中间项的两倍,即,:,a,1,+,a,n,=,a,2,+,a,n,-,1,=,a,3,+a,n,-,2,=,=2,a,中,.,a,1,+,a,n,=,a,2,+,a,n,-,1,=,a,3,+,a,n,-,2,=,.,二、等差数列的性质 1.首尾项性质: 有穷等差数,特别地,若,m,+,n,=2,p,则,a,m,+,a,n,=2,a,p,.,2.,若,p,+,q,=,r,+,s,(,p,、,q,、,r,、,s,N,*,),则,a,p,+,a,q,=,a,r,+,a,s,.,3.,等差中项,如果在两个数,a,、,b,中间插入一个数,A,使,a,、,A,、,b,成等差差数列,则,A,叫做,a,与,b,的等差中项,.,4.,顺次,n,项和性质,5.,已知,a,n,是公差为,d,的等差数列,a,+,b,A= .,2,(1),若,n,为奇数,则,S,n,=,na,中,且,S,奇,-,S,偶,=,a,中, = .,S,奇,S,偶,n,+1,n,-,1,(2),若,n,为偶数,则,S,偶,-,S,奇,=,.,nd,2,若,a,n,是公差为,d,的等差数列,则,a,k, ,a,k, ,a,k,也成等差数列,且公差为,n,2,d,.,k,=2,n,+1,3,n,k,=1,n,k,=,n,+1,2,n,特别地, 若 m+n=2p, 则 am+an=2ap,6.,若,a,n, ,b,n,均为等差数列,则,ma,n, ,ma,n,kb,n,也为等差数列,其中,m,k,均为常数,.,7.,若等差数列,a,n,的前,2,n,-,1,项和为,S,2,n,-,1,等差数列,b,n,的前,2,n,-,1,项和为,T,2,n,-,1,则,= .,S,2,n,-,1,T,2,n,-,1,a,n,b,n,例：已知等差数列 ， 的前,n,项的分别为 ， ，若,则,。,6.若 an, bn 均为等差数列,三、判断、证明方法,1.,定义法,;,2.,通项公式法,;,3.,等差中项法,.,四、,S,n,的最值问题,二次函数,注,:,三个数成等差数列,可设为,a,-,d,a,a,+,d,(,或,a,a,+,d,a,+2,d,),四个数成等差数列,可设为,a,-,3,d,a,-,d,a,+,d,a,+3,d,.,1.,若,a,1,0,d,0,时,满足,a,n,0,a,n,+1,0.,2.,若,a,1,0,时,满足,a,n,0,a,n,+1,0.,三、判断、证明方法1.定义法;2.通项公式法;3.等差中项法,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,没什么感觉,怎么办,?,那就要回到最基本的地方去,用首项和公差来分析,有用吗,?,需要尝试,!,没什么感觉,怎么办?那就要回到最基本的地方去,用首项和公差来,漂亮,!,!,漂亮!,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,典型例题,解,:,不妨设,Q,P,则,S,Q,-,S,P,=,a,P,+1,+,+,a,Q,=,-,.,P,+,Q,PQ,a,P,+1,+,a,Q,2,则,S,P,+,Q,= =,(,P,+,Q,)(,a,1,+,a,P,+,Q,),2,(,P,+,Q,)(,a,P,+1,+,a,Q,),2,(,P,+,Q,),2,PQ,=,-,.,1.,已知, ,成等差数列,求证,: , ,成等差数列,.,b,1,a,1,c,1,c,a,+,b,b,c,+,a,a,b,+,c,2.,等差数列的前,n,项和为,S,n,若,S,P,= ,S,Q,= (,P,Q,),求,S,P,+,Q,(,用,P,Q,表示,),.,Q,P,P,Q,3.,等差数列的前,n,项和为,S,n,若,S,m,=,S,k,(,m,k,),求,S,m,+,k,.,4.,等差数列,a,n,的首项,a,1,0,前,n,项和为,S,n,若,S,m,=,S,k,m,k,问,n,为何值时,S,n,最大,.,0,n,= (,m,+,k,为偶数时,);,或,(,m,+,k,为奇数时,).,m,+,k,2,m,+,k,+1,2,m,+,k,-,1,2,典型例题解: 不妨设 QP, 则 SQ-SP=aP+1+,5.,在等差数列,a,n,中,已知,a,1,=20,前,n,项和为,S,n,且,S,10,=,S,15,. (1),求前,n,项和,S,n,;,(2),当,n,为何值时,S,n,有最大值,并求它的最大值,.,(1),S,n,=,-,(,n,2,-,25,n,);,5,6,(2),当且,仅当,n,=12,或,13,时,S,n,有最大值,最大值为,130.,6.,已知等差数列,a,n,的前,n,项和为,S,n,且,a,2,=1,S,11,=33. (1),求数列,a,n,的通项公式,; (2),设,b,n,=( ) ,且数列,b,n,的前,n,项和为,T,n,求证,:,数列,b,n,是等比数列,并求,T,n,.,a,n,1,2,(1),a,n,=,n,;,1,2,(2),T,n,=(,2,+1)(1,-,2,-,).,n,2,5.在等差数列 an 中, 已知 a1=20,7.,已知函数,f,(,t,),对任意实数,x,y,都有,:,f,(,x,+,y,)=,f,(,x,)+,f,(,y,)+3,xy,(,x,+,y,+2)+3,f,(1)=1. (1),若,t,为正整数,试求,f,(,t,),的表达式,; (2),满 足,f,(,t,)=,t,的所有整数,t,能否构成等差数列,?,若能构成等差数列,求出此数列,;,若不能构成等差数列,请说明理由,; (3),若,t,为自然数,且,t,4,f,(,t,),mt,2,+(4,m,+1),t,+3,m,恒成立,求,m,的最大值,.,(1),f,(,t,)=,t,3,+3,t,2,-,3,(,t,N,*,);,(3),f,(,t,),mt,2,+(4,m,+1),t,+3,m,f,(,t,),-,t,m,(,t,2,+4,t,+3),m,t,-,1.,所求数列为,:,-,3,-,1, 1,或,1,-,1,-,3;,(2),f,(,t,)=,t,3,+3,t,2,-,3,(,t,Z,),f,(,t,)=,t,t,=,-,3,-,1, 1,故,m,的最大值是,3.,7.已知函数 f(t) 对任意实数 x, y 都有:,8.,已知函数,f,(,x,)=,px,2,+,qx,其中,p,0,p,+,q,1.,对于数列,a,n,设它的前项和为,S,n,且,S,n,=,f,(,n,),(,n,N,*,),. (1),求数列,a,n,的通项 公式,; (2),证明,:,a,n,+1,a,n,1; (3),证明,:,点,M,1,(1,),M,2,(2, ),M,3,(3,),M,n,(,n, ),都在同一直线上,.,1,S,1,2,S,2,3,S,3,n,S,n,(1),a,n,=(2,n,-,1),p,+,q,(,n,N,*,);,(2),a,n,+1,-,a,n,=2,p,0,a,n,+1,a,n,a,1,=,p,+,q,=1;,(3),只要证其中任意一点,M,r,(,r,)(,r,1,r,N,*,),与点,M,1,(1,),1,S,1,r,S,r,连线的斜率为定值,(,p,),即可,.,8.已知函数 f(x)=px2+qx, 其中,1.,已知,a,n,是等差数列,. (1),前,4,项和为,21,末,4,项和为,67,且各项和为,286.,求项数,; (2),S,n,=20,S,2,n,=38,求,S,3,n,; (3),项数为奇数,奇数项和为,44,偶数项和为,33,求数列的中间项和项数,.,解,:,(1),设数列的项数为,n,依题意得,:,4(,a,1,+,a,n,)=21+67=88.,a,1,+,a,n,=22.,由,n,(,a,1,+,a,n,)=2,S,n,=2,286,得,:,(2),S,n,S,2,n,-,S,n,S,3,n,-,S,2,n,成等差数列,S,3,n,-,S,2,n,+,S,n,=2(,S,2,n,-,S,n,).,a,1,+,a,2,+,a,3,+,a,4,=21,a,n,-,3,+,a,n,-,2,+,a,n,-,1,+,a,n,=67,且有,:,S,n,=286,a,1,+,a,n,=,a,2,+,a,n,-,1,=,a,3,+,a,n,-,2,=,a,4,+,a,n,-,3,.,n,=26.,故所求数列的项数为,26.,S,3,n,=3(,S,2,n,-,S,n,)=3(38,-,20)=54.,(3),依题意,S,奇,+,S,偶,=,S,n,S,奇,-,S,偶,=,a,中,S,n,=,na,中,.,S,n,=77,a,中,=11,S,n,=,na,中,.,解得,:,a,中,=11,n,=7.,课后练习题,1.已知 an 是等差数列. (1)前 4 项,2.,等差数列,a,n,b,n,中,前,n,项和分别为,S,n,S,n,且,=,求,.,S,n,S,n,7,n,+2,n,+4,a,5,b,5,解,:,a,n, ,b,n,是等差数列,它们的前,n,项和是关于,n,的二次函数,且常数项为,0,a,5,=,S,5,-,S,4,=65,k,b,5,=,S,5,-,S,4,=13,k,.,a,5,b,5,= =5.,65,k,13,k,S,9,S,9,7,9,+2,9+4,a,5,b,5,或,= = = = = =5.,a,1,+,a,9,2,b,1,+,b,9,2,a,1,+,a,9,2,b,1,+,b,9,2,9,9,13,65,可设,S,n,=,kn,(7,n,+2),S,n,=,kn,(,n,+4),2.等差数列 an, bn 中, 前 n 项,3.,设,a,n,是一个公差为,d,(,d,0),的等差数列,它的前,10,项和,S,10,=110,且,a,1,a,2,a,4,成等比数列,. (1),证明,:,a,1,=,d,; (2),求公差,d,的值和数列,a,n,的通项公式,.,(1),证,:,a,1,a,2,a,4,成等比数列,a,2,2,=,a,1,a,4,.,而,a,n,是等差数列,有,a,2,=,a,1,+,d,a,4,=,a,1,+3,d,.,(,a,1,+,d,),2,=,a,1,(,a,1,+3,d,),整理得,d,2,=,a,1,d,.,d,0,a,1,=,d,.,(2),解,:,S,10,=110,而,S,10,=10,a,1,+45,d,10,a,1,+45,d,=110,又由,(1),知,a,1,=,d,代入上式得,: 11,a,1,=22.,即,2,a,1,+9,d,=22.,a,1,=2.,a,n,=2+(,n,-,1),2=2,n,.,d,=,a,1,=2.,公差,d,的值为,2,数列,a,n,的通项公式为,a,n,=2,n,.,3.设 an 是一个公差为 d(d0) 的,4.,已知数列,a,n,满足,a,1,=4,a,n,=4,-,(,n,2),令,b,n,=,. (1),求证,:,数列,b,n,是等差数列,; (2),求数列,a,n,的通项公式,.,a,n,-,1,4,a,n,-,2,1,(1),证,:,由已知,a,n,+1,-,2=2,-,= .,4,a,n,2(,a,n,-,2),a,n,a,n,+1,-,2,1,= = + .,2(,a,n,-,2),a,n,a,n,-,2,1,1,2,-,= .,a,n,+1,-,2,1,a,n,-,2,1,1,2,即,b,n,+1,-,b,n,= .,1,2,故数列,b,n,是等差数列,.,(2),解,:, ,是等差数列,a,n,-,2,1,= +(,n,-,1),=,.,a,1,-,2,1,a,n,-,2,1,n,2,1,2,数列,a,n,的通项公式为,a,n,=2+,.,2,n,a,n,=2+ .,2,n,4.已知数列 an 满足 a1=4, an=4-,5.,数列,a,n,的前,n,项和为,S,n,=,npa,n,(,n,N,*,),且,a,1,a,2, (1),求常数,p,的值,;,(2),证明数列,a,n,是等差数列,.,(1),解,:,当,n,=1,时,a,1,=,pa,1,若,p,=1,则,当,n,=2,时有,a,1,+,a,2,=2,pa,2,=2,a,2,.,a,1,=,a,2,与,a,1,a,2,矛盾,.,p,1.,a,1,=0.,由,a,1,+,a,2,=2,pa,2,知,: (2,p,-,1),a,2,=,a,1,=0.,a,2,a,1,a,2,0,p,= .,1,2,(2),证,:,由已知,S,n,=,na,n,a,1,=0.,1,2,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=,na,n,-,(,n,-,1),a,n,-,1,1,2,1,2,= .,a,n,-,1,a,n,n,-,1,n,-,2,则,= , = .,a,n,-,2,a,n,-,1,n,-,2,n,-,3,a,2,a,3,2,1,=,n,-,1.,a,2,a,n,a,n,=(,n,-,1),a,2,.,a,n,-,a,n,-,1,=,a,2,.,故,数列,a,n,是以,a,1,为首项,a,2,为公差的等差数列,.,5.数列 an 的前 n 项和为 Sn=np,6.,已知,数列,a,n,a,n,N,*,S,n,= (,a,n,+2),2, (1),求证,:,a,n,是等差数列,; (2),若,b,n,=,a,n,-,30,求数列,b,n,的前,n,项和的最小值,.,1,2,1,8,(1),证,:,由,a,n,+1,=,S,n,+1,-,S,n,得,: 8,a,n,+1,=(,a,n,+1,+2),2,-,(,a,n,+2),2,.,(2),解,:,由已知,8,a,1,=8,S,1,=(,a,1,+2),2,a,1,=2,故,由,(1),知,a,n,=4,n,-,2.,(,a,n,+1,-,2),2,-,(,a,n,+2),2,=0.,(,a,n,+1,+,a,n,)(,a,n,+1,-,a,n,-,4)=0.,a,n,N,*,a,n,+1,+,a,n,0.,a,n,+1,-,a,n,-,4=0,即,a,n,+1,-,a,n,=4.,a,n,是等差数列,.,b,n,=2,n,-,1,-,30=2,n,-,31.,解,2,n,-,310,且,2(,n,+1),-,31,0,得,:,n,0,d,a,2,a,k,0,a,k,+1,.,由,a,k,=2,-,(,k,-,1),0,得,k,19.,1,9,由,k,=2,n,19(,n,N,*,),得,n,4.,即在数列,a,2,n,中,a,2,1,a,2,2,a,2,3,a,2,4,0,a,2,5,.,当,n,=4,时, A,n,的值最大,其最大值为,:,A,n,max,=,(19,4+2,-,2,4+1,)= .,1,9,9,46,7.已知等差数列 an 的首项是 2, 前 1,解,:,求,A,n,的最大值有以下解法,:,法,2,:,若存在,n,N,*,使得,A,n,A,n,+1,且,A,n,A,n,-,1,则,A,n,的值最大,.,=,(19,n,+2,-,2,n,+1,),1,9,A,n,A,n,A,n,+1,A,n,A,n,-,1,19,n,+2,-,2,n,+1,19(,n,+1)+2,-,2,n,+2,19,n,+2,-,2,n,+1,19(,n,-,1)+2,-,2,n,解得,: 9.5,2,n,19(,n,N,*,),n,=4,.,故,取,n,=4,时, A,n,的值最大,其最大值为,:,A,n,max,=,(19,4+2,-,2,4+1,)= .,1,9,9,46,7.,已知等差数列,a,n,的首项是,2,前,10,项之和是,15,记,A,n,=,a,2,+,a,4,+,a,8,+,+,a,2,n,(,n,N,*,),求,A,n,及,A,n,的最大值,.,解: 求 An 的最大值有以下解法:法2: 若存在 n,8.,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,:,设等差数列,a,n,的公差为,d,则,S,n,=,na,1,+,.,n,(,n,-,1),d,2,S,7,=7,S,15,=75,解得,:,a,1,=,-,2,d,=1.,T,n,=,n,2,-,n,.,9,4,1,4,7,a,1,+21,d,=7,15,a,1,+105,d,=75,a,1,+3,d,=1,a,1,+7,d,=5,即,=,a,1,+ (,n,-,1),d,=,-,2+ (,n,-,1).,S,n,n,1,2,1,2,-,= ,S,n,+1,n,+1,S,n,n,1,2,1,2,S,n,n,数列, ,是等差数列,其首项为,-,2,公差为,.,8.设 an 为等差数列, Sn 为数列 a,9.,两个数列,a,n,和,b,n,满足,b,n,=,求证,: (1),若,b,n,为等差数列,则数列,a,n,也是等差数列,;,(2),(1),的逆命题也成立,.,1+2+,+,n,a,1,+2,a,2,+,+,na,n,证,:,(1),由已知得,a,1,+2,a,2,+,+,na,n,=,n,(,n,+1),b,n,. ,1,2,a,1,+2,a,2,+,+,na,n,+(,n,+1),a,n,+1,= (,n,+1)(,n,+2),b,n,+1,. ,1,2,将,式减,式化简得,:,a,n,+1,=,(,n,+2),b,n,+1,-,nb,n,.,1,2,1,2,a,n,=,(,n,+1),b,n,-,(,n,-,1),b,n,-,1,=,(,n,+1),b,n,-,(,n,-,1)(2,b,n,-,b,n,+1,).,1,2,1,2,1,2,1,2,b,n,为等差数列,b,n,-,1,=2,b,n,-,b,n,+1,b,n,+1,-,b,n,为常数,.,a,n,+1,-,a,n,=,(,n,+2),b,n,+1,-,nb,n,-,(,n,+1),b,n,+,(,n,-,1)(2,b,n,-,b,n,+1,),1,2,1,2,1,2,1,2,=,(,b,n,+1,-,b,n,),为常数,.,3,2,故数列,a,n,也是等差数列,.,9.两个数列 an 和 bn 满足 bn,证,:,(2),(1),的逆命题为,:,两个数列,a,n,和,b,n,满足,:,1+2+,+,n,a,1,+2,a,2,+,+,na,n,b,n,= ,若,a,n,为等差数列,则数列,b,n,也是等差数列,.,证明如下,:,a,n,是等差数列,可设,a,n,=,an,+,b,(,a,b,为常数,),.,na,n,=,an,2,+,bn,.,a,1,+2,a,2,+,+,na,n,=,a,(1,2,+2,2,+,+,n,2,)+,b,(1+2+,+,n,).,1+2+,+,n,a,1,+2,a,2,+,+,na,n,b,n,= =,an,(,n,+1)(2,n,+1)+,bn,(n+1),n,(,n,+1),1,2,1,2,1,6,1,3,=,a,(2,n,+1)+,b,.,b,n,+1,-,b,n,=,a,为常数,.,2,3,故数列,b,n,也是等差数列,.,证: (2) (1)的逆命题为: 两个数列 an 和 ,10.,已知数列,a,n,是等差数列,其前,n,项和为,S,n,a,3,=7,S,4,=24. (1),求数列,a,n,的通项公式,; (2),设,p,q,是正整数,且,p,q,证明,:,a,1,+2,d,=7,且,4,a,1,+6,d,=24.,解得,:,a,1,=3,d,=2.,a,n,=,a,1,+(,n,-,1),d,=3+2(,n,-,1)=2,n,+1.,S,p,+,q, (,S,2,p,+,S,2,q,).,1,2,故数列,a,n,的通项公式为,a,n,=2,n,+1.,(2),证,:,由,(1),知,a,n,=2,n,+1,S,n,=,n,2,+2,n,.,(1),解,:,设,等差数列,a,n,的公差为,d,依题意得,:,2,S,p,+,q,-,(,S,2,p,+,S,2,q,)=2(,p,+,q,),2,+2(,p,+,q,),-,(4,p,2,+4,p,),-,(4,q,2,+4,q,),=,-,2(,p,-,q,),2,.,又,p,q,2,S,p,+,q,-,(,S,2,p,+,S,2,q,)0.,S,p,+,q, (,S,2,p,+,S,2,q,).,1,2,故,10.已知数列 an 是等差数列, 其前 n,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,高考链接：,高考链接：,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,省内模拟考链接,省内模拟考链接,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,人教版高中数学等差数列课件,