高三数学高考复习课本回扣训练 课件六 课件.ppt

,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,*,1.,已知数列,a,n,的前,n,项和,S,n,=,n,2,-9,n,(,n,N,*,),第,k,项满,足,5,a,k,8,则,k,等于,(),A.9 B.8 C.7 D.6,解析,因,a,1,=,S,1,=-8,而当,n,2,时,由,a,n,=,S,n,-,S,n,-1,求得,a,n,=,2,n,-10,此式对于,n,=1,也成立,.,要满足,5,a,k,8,只须,5,2,k,-10,8,从而有 而,k,为自然数,.,因而,只能取,k,=8.,回扣练习六,B,2.,如果,a,1,a,2,a,8,为各项都大于零的等差数列,公差,d,0,则,(),A.,a,1,a,8,a,4,a,5,B.,a,1,a,8,a,4,a,5,C.,a,1,+,a,8,a,4,+,a,5,D.,a,1,a,8,=,a,4,a,5,解析,a,8,=,a,1,+7,d,a,4,=,a,1,+3,d,a,5,=,a,1,+4,d,a,1,a,8,=,+,a,1,7,d,=+7,a,1,d,a,4,a,5,=(,a,1,+3,d,)(,a,1,+4,d,)=+7,a,1,d,+12,d,2,.,又,d,0,d,2,0,a,1,a,8,a,4,a,5,.,3.,已知数列,a,n,对于任意的,p,q,N,*,满足,a,p,+,q,=,a,p,+,a,q,且,a,2,=-6,则,a,10,等于,(),A.-165 B.-33 C.-30 D.-21,解析,依题意知,:,a,10,=,a,2+8,=,a,2,+,a,8,=,a,2,+,a,4,+,a,4,=5,a,2,=-30.,B,C,4.,在正项等比数列,a,n,中,若,a,2,a,4,a,6,a,8,a,10,=32,等于,(),A.B.C.D.,解析,由,a,2,a,4,a,6,a,8,a,10,=32,得,a,6,=2,有等比数列的性质得,:,D,5.,已知等比数列,a,1,a,2,a,3,的和为定值,m,(,m,0),且其,公比,q,0,令,t,=,a,1,a,2,a,3,则,t,的取值范围是,(),A.(0,m,3,B.-,m,3,0),C.-,m,3,m,3,D.-,m,3,0)(0,m,3,解析,因为,m,=,a,1,+,a,2,+,a,3,=,所以 因为,q,0,所以,(,当且仅当,q,=-1,时取等号,).,又,m,0,所以,故,t,=,a,1,a,2,a,3,=,B,6.,已知等差数列,a,n,中,a,n,0,若,m,1,且,a,m,-1,-,+,a,m,+1,=0,S,2,m,-1,=38,则,m,的值为,(),A.38 B.20 C.19 D.10,解析,由,a,n,为等差数列得,a,m,-1,+,a,m,+1,=2,a,m,(,m,1,m,N,*,),又这里,a,m,-1,-,+,a,m,+1,=0,故得,则这里,a,m,0,a,m,=2,，再由,S,2,m,-1,=38,得,得,2,m,-1=19,解得,m,=10.,D,7.,已知等比数列的公比为,2,且前,4,项之和等于,1,那么,它的前,8,项之和等于,_.,解析,由已知得,S,4,=1,q,=2,8.,设,a,n,是公比为,q,的等比数列,S,n,是它的前,n,项和,若,S,n,是等差数列,则,q,=_.,解析,注意到,又,S,n,为等差数列,当,n,2,时,a,n,=,S,n,-,S,n,-1,=,=,S,2,-,S,1,=,a,2,a,1,q,n,-1,=,a,1,q,而,a,1,0,q,n,-2,=1,即,q,=1.,17,1,9.,各项都是正数的等比数列,a,n,的公比,q,1,且,a,2,a,1,成等差数列,则,=_.,解析,注意到 只要求出,q,;,由已知条件得 ,a,1,q,2,=,a,1,(1+,q,),q,2,-,q,-1=0,由此解得,q,=,a,n,0,q,0,q,=,于是得,10.,设数列,a,n,的前,n,项和为,S,n,(,n,1),且,a,4,=54,则,a,1,=_.,解析,由已知得,a,4,=,S,4,-,S,3,=,a,1,(81-1)-,a,1,(27-1)=108,54,a,1,=108,a,1,=2.,11.,证明,:,若,f,(,x,)=,ax,+,b,且,x,n,是等差数列,则,f,(,x,n,),也,是等差数列,.,证明,已知,x,n,是等差数列,设任意,n,N,，,有,x,n,+1,-,x,n,=,r,(,r,是公差,).,于是,任意,n,N,有,f,(,x,n,+1,)-,f,(,x,n,)=(,ax,n,+1,+,b,)-(,ax,n,+,b,),=,a,(,x,n,+1,-,x,n,)=,ar,其中,ar,是常数,即,f,(,x,n,),也是等差数,列,.,2,12.,已知数列,a,n,的前,n,项和,S,n,=,n,2,-48,n,.,(1),求数列的通项公式；,(2),求,S,n,的最大或最小值,.,解,(1),a,1,=,S,1,=1,2,-48,1=-47,当,n,2,时,a,n,=,S,n,-,S,n,-1,=,n,2,-48,n,-(,n,-1),2,-48(,n,-1),=2,n,-49,a,1,也适合上式，,a,n,=2,n,-49(,n,N,+,).,(2),a,1,=-49,d,=2,所以,S,n,有最小值,n,=24,即,S,n,最小,或,:,由,S,n,=,n,2,-48,n,=(,n,-24),2,-576,当,n,=24,时,S,n,取得最小值,-576.,返回,