高中数学 第一章 数列 1.2 等差数列 1.2.2.2 等差数列前n项和的性质课件 北师大版必修5

-,*,-,*,第,2,课时等差数列前,n,项和的性质,1,1,.,掌握等差数列与其前,n,项和,S,n,有关的性质,并能熟练运用这些性质解题,.,2,.,掌握等差数列与函数之间的关系,.,2,1,.,等差数列前,n,项和的性质,(1),在等差数列,a,n,中,每,m,项的和,a,1,+a,2,+,+a,m,a,m+,1,+a,m+,2,+,+a,2,m,a,2,m+,1,+a,2,m+,2,+,+a,3,m,仍为等差数列,即,S,m,S,2,m,-S,m,S,3,m,-S,2,m,仍为等差数列,.,(2),在等差数列,a,n,中,公差为,d,S,奇,表示奇数项的和,S,偶,表示偶数项的和,当,n,为奇数时,3,【做一做,1,】,已知等差数列,a,n,共有,10,项,其奇数项之和为,15,偶数项之和为,30,则其公差,d,为,(,),.,A.5B.4C.3D.2,解析,:,解法一,:,由题意得,S,偶,-S,奇,=,5,d=,15,所以,d=,3,.,答案,:,C,4,2,.,等差数列,a,n,的前,n,项和,S,n,的最值,首先要明确,此类问题一定是针对变号的等差数列而言,.,因为当公差,d,0,时,等差数列一定是单调数列,所以只有一个变号点,.,常用方法如下,:,(1),定义法,:,S,n,=a,1,+a,2,+,+a,n,.,从等差数列的单调性分析,S,n,值的变化,.,当,a,1,0,d,0,时,S,n-,1,S,n,递增,;,当,a,n+,1,S,n+,1,递减,;,类似地,当,a,1,0,时,n,为使,a,n,0,成立的最大自然数时,S,n,最小,.,5,【做一做,2,-,1,】,已知等差数列,a,n,中,|a,3,|=|a,9,|,公差,d,0,则使前,n,项和,S,n,取得最大值的自然数,n,是,.,解析,:,|a,3,|=|a,9,|,且,d,0,a,3,=-a,9,.,a,3,+a,9,=,0,即,a,6,=,0,.,使前,n,项和,S,n,取得最大值的自然数,n,是,5,或,6,.,答案,:,5,或,6,【做一做,2,-,2,】,在等差数列,a,n,中,若,a,1,=-,11,a,4,=-,5,则当前,n,项和,S,n,取最小值时,n=,.,5,.,5,n,6,.,5,.,又,n,N,+,n=,6,.,答案,:,6,6,题型一,题型二,题型三,题型一,等差数列前,n,项和的性质,【例,1,】,(1),项数为奇数的等差数列,奇数项之和为,44,偶数项之和为,33,求这个数列的中间项及项数,.,(2),一个等差数列的前,10,项之和为,100,前,100,项之和为,10,求前,110,项之和,.,分析,:(1),考查等差数列奇数项、偶数项、特殊项和项数之间的关系,.,(2),本题基本解法是求,a,1,d,或令,S,n,=an,2,+bn,求,S,n,再求,S,110,或利用性质,.,7,题型一,题型二,题型三,解,:,(1),设等差数列,a,n,共有,2,n-,1,项,则奇数项有,n,项,偶数项共有,n-,1,项,中间项为,a,n,.,S,奇,-S,偶,=a,n,=,44,-,33,a,n,=,11,.,这个数列,a,n,的中间项为,11,项数为,7,.,8,题型一,题型二,题型三,9,题型一,题型二,题型三,10,题型一,题型二,题型三,11,题型一,题型二,题型三,反思,若,a,n,是等差数列,S,n,是前,n,项和,则,(1),S,n,S,2,n,-S,n,S,3,n,-S,2,n,成等差数列,;,12,题型一,题型二,题型三,【变式训练,1,】,已知等差数列,a,n,共有,12,项,且前,12,项的和为,144,其中偶数项的和比奇数项的和大,12,求,S,奇,、,S,偶,及公差,d.,13,题型一,题型二,题型三,题型二,等差数列前,n,项和的最值问题,【例,2,】,在等差数列,a,n,中,a,1,=,25,S,17,=S,9,求前,n,项和,S,n,的最大值,.,分析,:,建立,S,n,关于,n,的二次函数式,利用二次函数求最大值,;,也可确定,a,n,0,a,n+,1,0,时,n,的值,从而确定,S,n,的最大值,.,14,题型一,题型二,题型三,解法三,:,先求出,d=-,2(,同解法一,),.,由,S,17,=S,9,得,a,10,+a,11,+,+a,17,=,0,.,而,a,10,+a,17,=a,11,+a,16,=a,12,+a,15,=a,13,+a,14,故,a,13,+a,14,=,0,.,d=-,2,0,a,13,0,a,14,0,S,13,0,S,13,0,17,题型一,题型二,题型三,18,题型一,题型二,题型三,题型三,易错辨析,易错点,:,对等差数列前,n,项和的性质理解不清致误,【例,3,】,等差数列,a,n,的前,m,项的和为,30,前,2,m,项的和为,100,则它的前,3,m,项的和为,(,),.,A.130B.170C.210D.260,错解,:A,或,B,错因分析,:,本题容易错选,A,或,B,.,由数列,a,n,是等差数列,错误地认为,S,3,m,=S,m,+S,2,m,从而得出,S,3,m,=,130,故错选,A,.,由数列,a,n,是等差数列,错误地认为,S,m,S,2,m,S,3,m,仍成等差数列,从而得出,S,3,m,=,170,故错选,B,.S,m,S,2,m,-S,m,S,3,m,-S,2,m,成等差数列,而不是,S,m,S,2,m,S,3,m,成等差数列,.,19,题型一,题型二,题型三,正解,:,解法一,:,取特殊值,m=,1,S,m,=S,1,=a,1,=,30,S,2,m,=S,2,=a,1,+a,2,=,100,则,S,3,m,=S,3,=a,1,+a,2,+a,3,=,3,a,2,=,3(,S,2,-S,1,),=,210,.,解法二,:,由题意可知,S,m,S,2,m,-S,m,S,3,m,-S,2,m,成等差数列,S,3,m,-S,2,m,=,(,S,2,m,-S,m,),2,-S,m,=,(100,-,30),2,-,30,=,110,.,S,3,m,=S,2,m,+,110,=,100,+,110,=,210,.,答案,:,C,20,1,2,3,4,A.5B.6,C.5,或,6D.6,或,7,该数列是等差数列,公差,d,0,a,1,-a,11,0,.,a,1,+a,11,=,0,.,a,6,=,0,.,S,5,或,S,6,最大,.,答案,:,C,21,1,2,3,4,2,在等差数列,a,n,中,S,n,是其前,n,项和,且,S,2 011,=S,2 015,S,k,=S,2 009,则正整数,k,为,(,),.,A.2 014B.2 015,C.2 016D.2 017,解析,:,因为等差数列的前,n,项和,S,n,可看成是关于,n,的二次函数,所以由二次函数的对称性及,S,2,011,=S,2,015,S,k,=S,2,009,答案,:,D,22,1,2,3,4,3,设等差数列,a,n,的前,n,项和为,S,n,.,若,a,2,=-,9,a,4,+a,6,=-,6,则当,S,n,取最小值时,n,等于,(,),.,A.9B.8,C.7D.6,答案,:,D,23,1,2,3,4,4,设等差数列,a,n,的前,n,项和为,S,n,且,S,m,=-,2,S,m+,1,=,0,S,m+,2,=,3,则,m=,.,答案,:,4,24,