人教版《等比数列》完美推荐1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,2020/10/27,#,3,.,2,等比数列的前,n,项和,1,.,等比数列的前,n,项和公式,若,数列,a,n,是公比为,q,的等比数列,则,当,q=,1,时,S,n,=,na,1,;,【,做一做,1】,已知,等比数列,a,n,的首项为,2,公比为,-,1,则数列,a,n,的前,99,项之和是,.,答案,:,2,2,.,等比数列前,n,项和的性质,(1),S,n+m,=S,n,+q,n,S,m,;,(2),在等比数列中,若项数为,2,n,(,n,N,+,),则,(3),当,q=-,1,且,k,为偶数时,S,k,S,2,k,-S,k,S,3,k,-S,2,k,不是等比数列,;,当,q,-,1,或,k,为奇数时,S,k,S,2,k,-S,k,S,3,k,-S,2,k,是等比数列,.,【,做一做,2,】,解析,:,根据等比数列的性质,S,3,S,6,-S,3,S,9,-S,6,仍然成等比数列,.,答案,:,B,知识拓展,等比数列前,n,项和公式的推导,推导前,n,项和公式的方法,除了教材上提供的错位相减法,还有以下几种方法,.,当,q=,1,时,数列,a,n,变为,a,1,a,1,a,1,a,1,易得它的前,n,项和,S,n,=na,1,.,(2)(,拆项法,):,由,S,n,=a,1,+a,2,+a,3,+,+a,n,=a,1,+q,(,a,1,+a,2,+a,3,+,+a,n-,1,),=a,1,+qS,n-,1,=a,1,+q,(,S,n,-a,n,),得,(1,-q,),S,n,=a,1,-a,n,q.,结论同上,.,(3)(,累加法,):,a,n,=a,1,+,(,a,2,-a,1,),+,(,a,3,-a,2,),+,+,(,a,n,-a,n-,1,),=a,1,+a,1,(,q-,1),+a,2,(,q-,1),+,+a,n-,1,(,q-,1),=a,1,+,(,a,1,+a,2,+,+a,n-,1,)(,q-,1),=a,1,+,(,S,n,-a,n,)(,q-,1),整理,得,(1,-q,),S,n,=a,1,-a,n,q.,结论同上,.,3,.,乘公比错位相减法求和,课本上推导等比数列前,n,项和的方法,即错位相减法,它解决的主要求和问题是,:,如果数列,a,n,和,b,n,分别是等差和等比数列,求数列,a,n,b,n,的前,n,项和,.,求和过程如下,:,设数列,a,n,b,n,的前,n,项和是,S,n,等差数列,a,n,的公差是,d,等比数列,b,n,的公比是,q,则有,S,n,=a,1,b,1,+a,2,b,2,+a,3,b,3,+,+a,n,b,n,=a,1,b,1,+a,2,b,1,q+a,3,b,1,q,2,+,+a,n,b,1,q,n-,1,qS,n,=a,1,b,1,q+a,2,b,1,q,2,+a,3,b,1,q,3,+,+a,n,b,1,q,n,S,n,-qS,n,=a,1,b,1,+,(,a,2,-a,1,),b,1,q+,(,a,3,-a,2,),b,1,q,2,+,+,(,a,n,-a,n-,1,),b,1,q,n-,1,-a,n,b,1,q,n,.,【,做一做,3】32,-,1,+,42,-,2,+,52,-,3,+,+,(,n+,2)2,-n,=,.,思考辨析,判断下列说法是否正确,正确的在后面的括号内打,“,”,错误的打,“,”,.,(1),若数列,a,n,是等比数列,S,n,是数列,a,n,的前,n,项和,则,S,n,S,2,n,-S,n,S,3,n,-S,2,n,一定成等比数列,.,(,),(2),数列,a,a,2,a,3,a,n,的前,n,项和为,.,(,),(3),若等比数列,a,n,的前,n,项和为,S,n,=,3,n,+t,则,t=-,1,.,(,),答案,:,(1),(2),(3),探究一,探究二,探究三,思维辨析,【,例,1,】,(1),a,n,为等比数列,若,a,1,+a,3,=,10,a,4,+a,6,=,求,a,4,和,S,5,;,(2),在等比数列,a,n,中,a,1,+a,n,=,66,a,2,a,n-,1,=,128,其前,n,项和为,S,n,S,n,=,126,求,n,和,q.,探究一,探究二,探究三,思维辨析,探究一,探究二,探究三,思维辨析,反思感悟,1,.,在等比数列的前,n,项和公式中,共有,a,1,a,n,q,n,S,n,这五个量,已知其中任何三个量,都可以求其余两个量,.,2,.,求解等比数列的计算问题,多采用基本量方法,即建立关于,a,1,和,q,的方程,(,组,),求得,a,1,与,q,后再解决其他问题,.,3,.,应用等比数列前,n,项和公式时,必须注意,q=,1,与,q,1,这两种情况,.,4,.,在等比数列的前,n,项和,S,n,中,当,n,值较小时,可直接用,a,1,+a,2,+,+a,n,来表示,S,n,如,S,3,=a,1,+a,2,+a,3,=a,1,+a,1,q+a,1,q,2,.,探究一,探究二,探究三,思维辨析,变式训练,1,探究一,探究二,探究三,思维辨析,【,例,2,】,(1),等比数列,a,n,共有,2,n,项,其和为,-,240,且奇数项的和比偶数项的和大,80,则公比,q=,;,(2),在等比数列,a,n,中,S,2,=,7,S,6,=,91,则,S,4,=,.,分析,:,(1),先分别求出奇数项的和与偶数项的和,再运用性质求解,;(2),根据,S,2,S,4,-S,2,S,6,-S,4,成等比数列求解,.,探究一,探究二,探究三,思维辨析,(2),由等比数列前,n,项和性质,得,S,2,S,4,-S,2,S,6,-S,4,成等比数列,有,(,S,4,-S,2,),2,=S,2,(,S,6,-S,4,),即,(,S,4,-,7),2,=,7,(91,-S,4,),解得,S,4,=,28(,负值舍去,),.,答案,:,(1)2,(2)28,探究一,探究二,探究三,思维辨析,反思感悟,应用等比数列前,n,项和的性质可以避免烦琐的计算,使解题过程简化,常用的前,n,项和的性质是,:,(,2),在等比数列,a,n,中,若,S,m,S,2,m,-S,m,S,3,m,-S,2,m,均不为,0,则,S,m,S,2,m,-S,m,S,3,m,-S,2,m,成等比数列,.,运用此性质时,注意各项非零的要求及下标和差的构造,.,探究一,探究二,探究三,思维辨析,变式训练,2,已知等比数列前,n,项和为,S,n,若,S,2,=,4,S,4,=,16,则,S,8,=,(,),A.160B.64C.,-,64D.,-,160,解析,:,由等比数列的性质可得,S,2,S,4,-S,2,S,6,-S,4,S,8,-S,6,成等比数列,又,S,2,=,4,S,4,=,16,故,S,4,-S,2,=,12,所以公比为,3,由等比数列可得,S,6,-S,4,=,36,S,8,-S,6,=,108,解得,S,6,=,52,S,8,=,160,故选,A,.,答案,:,A,探究一,探究二,探究三,思维辨析,【,例,3,】,已知数列,a,n,的前,n,项和为,S,n,且,S,n,=,2,n,2,+n,n,N,+,数列,b,n,满足,a,n,=,4log,2,b,n,+,3,n,N,+,.,(1),求,a,n,b,n,;,(2),求数列,a,n,b,n,的前,n,项和,T,n,.,探究一,探究二,探究三,思维辨析,解,:,(1),由,S,n,=,2,n,2,+n,得当,n=,1,时,a,1,=S,1,=,3;,当,n,2,时,a,n,=S,n,-S,n-,1,=,4,n-,1,a,1,=,3,也符合上式,所以,a,n,=,4,n-,1,n,N,+,.,由,4,n-,1,=a,n,=,4log,2,b,n,+,3,得,b,n,=,2,n-,1,n,N,+,.,(2),由,(1),知,a,n,b,n,=,(4,n-,1)2,n-,1,n,N,+,所以,T,n,=,3,+,7,2,+,11,2,2,+,+,(4,n-,1)2,n-,1,2,T,n,=,3,2,+,7,2,2,+,+,(4,n-,5)2,n-,1,+,(4,n-,1)2,n,所以,2,T,n,-T,n,=,(4,n-,1)2,n,-,3,+,4,(2,+,2,2,+,+,2,n-,1,),=,(4,n-,5)2,n,+,5,.,故,T,n,=,(4,n-,5)2,n,+,5,n,N,+,.,探究一,探究二,探究三,思维辨析,反思感悟,1,.,一般地,如果数列,a,n,是等差数列,b,n,是等比数列且公比为,q,求数列,a,n,b,n,的前,n,项和时,可采用乘公比错位相减法,.,2,.,乘公比错位相减法的步骤如下,.,(1),作和,:,S,n,=a,1,b,1,+a,2,b,2,+,+a,n,b,n,.,(2),乘公比,:,qS,n,=a,1,b,2,+a,2,b,3,+,+a,n,b,n+,1,.,(3),相减,:(1,-q,),S,n,=a,1,b,1,+,(,a,2,-a,1,),b,2,+,+,(,a,n,-a,n-,1,),b,n,-a,n,b,n+,1,.,(4),化简求得,S,n,.,3,.,在写出,“,S,n,”,与,“,qS,n,”,的表达式时,应特别注意将两式,“,错项对齐,”,以便下一步准确写出,“,S,n,-qS,n,”,的表达式,并且要注意两式相减时式子的第一项和最后一项以及符号、项数等,.,探究一,探究二,探究三,思维辨析,变式训练,3,求数列,1,3,a,5,a,2,7,a,3,(2,n-,1,),a,n-,1,(,a0,),的前,n,项和,.,探究一,探究二,探究三,思维辨析,因忽视公式的应用条件而出错,典例,在等比数列,a,n,中,若,S,3,=,3,a,3,求其公比,q.,探究一,探究二,探究三,思维辨析,纠错心得,由于等比数列中公比,q,不确定,因此需要对,q,进行分类讨论,.,本题的错误就在于直接应用了,q,1,时的公式,虽然最后的结果是正确的,但解题过程错误,.,探究一,探究二,探究三,思维辨析,变式训练,等比数列,a,n,的前,n,项和为,S,n,若,5,S,1,=S,2,+S,3,且,S,4,=,10,则,a,n,=,.,1,2,3,4,5,1,.,等比数列,2,4,8,16,的前,n,项和为,(,),A.2,n+,1,-,1B.2,n,-,2C.2,n,D.2,n+,1,-,2,解析,:,由题意知,a,1,=,2,q=,2,则,答案,:,D,1,2,3,4,5,2,.,已知一个等比数列首项为,1,项数是偶数,其奇数项之和为,85,偶数项之和为,170,则这个数列的公比和项数分别为,(,),A.8,2B.2,4C.4,10D.2,8,解析,:,2,n,=,256,=,16,2,=,2,8,n=,8,.,答案,:,D,1,2,3,4,5,1,2,3,4,5,4,.,在等比数列,a,n,中,前,n,项和为,S,n,已知,S,10,=,48,S,20,=,60,则,S,30,=,.,解析,:,因为,a,n,是等比数列,所以,S,10,S,20,-S,10,S,30,-S,20,成等比数列,即,48,12,S,30,-,60,满足,12,2,=,48(,S,30,-,60),解得,S,30,=,63,.,答案,:,63,1,2,3,4,5,5,.,已知等比数列,a,n,的前,n,项和为,S,n,=a,2,n,+b,且,a,1,=,3,.,(1),求,a,b,的值及数列,a,n,的通项公式,;,(2),设,求数列,b,n,的前,n,项和,T,n,.,解,:,(1),当,n,2,时,a,n,=S,n,-S,n-,1,=,2,n-,1,a,而,a,n,为等比数列,所以,a,n,=,2,n-,1,a,所以,a,1,=,2,1,-,1,a=a=,3,从而,a,