高考数学 25个必考点 专题12 数列的通项优质课件

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,*,数列通项的求法,数列,高考数学,25,个必考点,专题复习策略指导,1,求,数列,通项公式,常用方法,2,、叠加法与累乘法求数列通项公式,3,、对于含递推关系的数列,构造出新的等差或等比数列来求通项,.,2,(,1,),当,n,2,时，,a,n,=S,n,S,n,1,=4,n,5;,当,n,=1,时，,a,1,=S,1,=,1+,k,;,当,k,=0,时，,a,1,=,1,适合,a,n,=4,n,5,a,n,=4,n,5;,当,k,0,时，,a,n,=,1+,k,不适合,a,n,=4,n,5,例,1,：已知数列,a,n,的前,n,项和为,S,n,，求,a,n,的通项,a,n,.,(,1,),S,n,=2,n,2,3,n,+,k,;,(2),a,n,=,1+,k,（,n,=1,）,4,n,5(,n,2).,解析,=2,n,2,3,n,+,k,2(,n,1),2,+3(,n,1),k,3,(2),a,n+1,=S,n+1,S,n,(,a,n+1,1),2,(,a,n,+1),2,=0,即,(,a,n+1,+,a,n,)(,a,n+1,a,n,2)=0,a,n,0,，,a,n+1,a,n,=2.,又,a,1,=1,，故,a,n,是首项为,1,，公差为,2,的等差数列，,a,n,=2n,1.,解析,例,1,：已知数列,a,n,的前,n,项和为,S,n,，求,a,n,的通项,a,n,.,(,1,),S,n,=2,n,2,3,n,+,k,;,(2),(,a,n+1,+1),2,4,a,n+1,(,a,n,+1),2,=0,4,例,1,：已知数列,a,n,的前,n,项和为,S,n,，求,a,n,的通项,a,n,.,(,1,),S,n,=2,n,2,3,n,+,k,;,(2),法二,5,叠加法与累乘法求数列通项公式,规律总结：,形如,叠加法,解析,6,解析,A,ln(,n,1),ln,n,.,即：,a,n,2,ln,n,.,7,规律总结：,形如,累乘法,解析,8,构造法求通项公式,解析,得：,c,=1.,a,n+1,+,1,=2(,a,n,+,1,),例,1:,在数列,a,n,中，,a,1,5,，,a,n,1,2,a,n,1,，则,a,n,=,；,9,解析,又,b,1,a,1,5.,变,1:,在数列,a,n,中，,a,1,5,，,a,n,1,2,a,n,2,n,，则,a,n,=,；,10,变,2:,在数列,a,n,中，,a,1,5,，,a,n,1,2,a,n,3,n,，则,a,n,=,；,解析,又,b,1,a,1,5.,2,n,+3,n,11,解析,总结：对于含递推关系的数列,构造出新的,等差或等比,数列来求通项,.,12,解析,(1),a,1,2,a,2,(,a,1,a,2,),4,，,a,1,2,a,2,3,a,3,2(,a,1,a,2,a,3,),6,，,13,证明,(2),即,S,n,2,S,n,1,2,，,S,n,1,20,，,14,see you!,15,