高三数学数列通项的求法1省名师优质课赛课获奖课件市赛课一等奖课件.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,*,本资料仅供参考，不能作为科学依据。谢谢。本资料仅供参考，不能作为科学依据。感谢,届高考数学复习,强化双基系列课件,1/18,34,数列通项求法,2/18,一、公式法,二、,迭,加法,若,a,n,+1,=,a,n,+,f,(,n,),则,:,若,a,n,+1,=,f,(,n,),a,n,则,:,三、,叠,乘法,a,n,=,S,1,(,n,=1),S,n,-,S,n,-,1,(,n,2).,a,n,=,a,1,+,(,a,k,-,a,k,-,1,)=,a,1,+,f,(,k,-,1,)=,a,1,+,f,(,k,).,n,-,1,k,=1,n,k,=2,n,k,=2,a,n,=,a,1,=,a,1,f,(1),f,(2),f,(,n,-,1)(,n,2).,a,n,a,n,-,1,a,2,a,1,a,3,a,2,3/18,四、化归法,经过恰当恒等变形,如配方、因式分解、取对数、取倒数等,转化为等比数列或等差数列,.,(1),若,a,n,+1,=,pa,n,+,q,则,:,a,n,+1,-,=,p,(,a,n,-,).,(3),若,a,n,+1,=,pa,n,+,q,(,n,),则,:,(2),若,a,n,+1,=,则,:,pa,n,r,+,qa,n,a,n,+1,1,a,n,1,=,+.,p,r,p,q,(4),若,a,n,+1,=,pa,n,q,则,:,lg,a,n,+1,=,q,lg,a,n,+lg,p,.,五、归纳法,先计算数列前若干项,经过观察规律,猜测通项公式,进而用数学归纳法证之,.,例,已知数列,a,n,满足,:,a,1,=1,a,n,+1,=2,a,n,+3,2,n,-,1,求,a,n,通项公式,.,a,n,=(3,n,-,1),2,n,-,2,a,n,+1,p,n,+1,a,n,p,n,=+,.,q,(,n,),p,n,+1,4/18,1.,在数列,a,n,中,a,1,=1,S,n,=(,n,2),求,a,n,.,S,n,-,1,2,S,n,-,1,+1,S,n,-,1,2,S,n,-,1,+1,解,:,由,S,n,=,知,:,1,S,n,1,S,n,-,1,-,=2.,1,S,n,是以,=1,为首项,公差为,2,等差数列,.,1,S,1,1,a,1,1,S,n,=1+2(,n,-,1)=2,n,-,1.,S,n,=.,2,n,-,1,1,a,1,=1,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=,-,.,(2,n,-,1)(2,n,-,3),2,a,n,=,-,n,2.,1,n,=1,(2,n,-,1)(2,n,-,3),2,经典例题,5/18,2.,数列,a,n,前,n,项和,S,n,=,n,2,-,7,n,-,8,(1),求,a,n,通项公式,;(2),求,|,a,n,|,前,n,项和,T,n,.,解,:,(1),当,n,=1,时,a,1,=,S,1,=,-,14;,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=2,n,-,8,(2),由,(1),知,当,n,4,时,a,n,0;,当,n,5,时,a,n,0;,当,n,5,时,T,n,=,-,S,4,+,S,n,-,S,4,=,S,n,-,2S,4,故,a,n,=,2,n,-,8,n,2.,-,14,n,=1,=,n,2,-,7,n,-,8,-,2(,-,20),当,n,4,时,T,n,=,-,S,n,=,-,n,2,+7,n,+8,=,n,2,-,7,n,+32.,故,T,n,=,n,2,-,7,n,+32,n,5.,-,n,2,+7,n,+8,n,4,6/18,3.,已知数列,a,n,中,a,1,=1,a,n,+1,=,a,n,+1(,n,N,*,),求,a,n,.,1,2,解法一,a,n,+1,=,a,n,+1(,n,N,*,),1,2,a,n,=,a,n,-,1,+1,a,n,-,1,=,a,n,-,2,+1.,1,2,1,2,两式相减得,:,a,n,-,a,n,-,1,=,(,a,n,-,1,-,a,n,-,2,),1,2,a,n,-,a,n,-,1,是以,a,2,-,a,1,=,为首项,公比为 等比数列,.,1,2,1,2,a,n,-,a,n,-,1,=,(,),n,-,2,=(,),n,-,1,.,1,2,1,2,1,2,a,n,=,a,1,+(,a,2,-,a,1,)+(,a,3,-,a,2,)+,+,(,a,n,-,a,n,-,1,),=1+,+(),2,+,+,(),n,-,1,1,2,1,2,1,2,=2,-,2,1,-,n,.,即,a,n,=2,-,2,1,-,n,.,7/18,解法二,由解法一知,a,n,-,a,n,-,1,=2,1,-,n,又,a,n,=,a,n,-,1,+1,1,2,消去,a,n,-,1,得,a,n,=2,-,2,1,-,n,.,解法三,a,n,=,a,n,-,1,+1,1,2,令,a,n,+,=,(a,n,-,1,+,),1,2,则,=,-,2.,a,n,-,2=,(,a,n,-,1,-,2).,1,2,a,n,-,2,是以,a,1,-,2=,-,1,为首项,公比为 等比数列,.,1,2,1,2,a,n,-,2=,-,(,),n,-,1,.,即,a,n,=2,-,2,1,-,n,.,3.,已知数列,a,n,中,a,1,=1,a,n,+1,=,a,n,+1(,n,N,*,),求,a,n,.,1,2,8/18,4.,数列,a,n,前,n,项和,S,n,满足条件,lg,S,n,+(,n,-,1)lg,b,=lg(,b,n,+1,+,n,-,2),其中,b,0,且,b,1.(1),求数列,a,n,通项公式,;(2),若对,n,N,*,n,4,时,恒有,a,n,+1,a,n,试求,b,取值范围,.,解,:,(1),由已知得,lg,S,n,b,n,-,1,=lg(,b,n,+1,+,n,-,2),S,n,b,n,-,1,=,b,n,+1,+,n,-,2(,b,1).,S,n,=,b,2,+(,b,1).,b,n,-,1,n,-,2,当,n,=1,时,a,1,=,S,1,=,b,2,-,1;,当,n,2,时,a,n,=,S,n,-,S,n,-,1,=,b,2,+,-,b,2,-,b,n,-,1,n,-,2,b,n,-,2,n,-,3,b,n,-,1,(1,-,b,),n,+3,b,-,2,=.,b,n,-,1,(1,-,b,),n,+3,b,-,2,n,2.,b,2,-,1,n,=1,故,a,n,=,(2),由已知,对,n,4,恒成立,.,b,n,-,1,(1,-,b,),n,+3,b,-,2,b,n,(1,-,b,)(,n,+1)+3,b,-,2,即,(,n,-,3),b,2,-,2(,n,-,2),b,+(,n,-,1)0,对,n,4,恒成立,.,亦即,(,b,-,1)(,n,-,3),b,-,(,n,-,1)0,对,n,4,恒成立,.,b,1,b,对,n,4,恒成立,.,n,-,3,n,-,1,n,-,3,n,-,1,而 当,n,=4,时有最大值,3,b,3.,9/18,5.,设,S,n,是等差数列,a,n,前,n,项和,.,已知,S,3,与,S,4,等比中项为,S,5,S,3,与,S,4,等差中项为,1,求等差数列,a,n,通项,a,n,.,1,5,1,3,1,4,1,3,1,4,解法,1:,设等差数列,a,n,首项,a,1,=,a,公差为,d,则通项公式为,a,n,=,a,+(,n,-,1),d,前,n,项和为,S,n,=,na,+.,n,(,n,-,1),d,2,1,3,1,4,依题意有,S,3,S,4,=(,S,5,),2,(,S,5,0),1,5,S,3,+,S,4,=2,1,3,1,4,由此可得,:,1,3,1,4,(3,a,+3,d,),(4,a,+6,d,)=,(5,a,+10,d,),2,1,4,(3,a,+3,d,),+,(4,a,+6,d,)=2.,1,3,25,1,整理得,3,ad,+5,d,2,=0,4,a,+5,d,=4.,解得,d,=0,a,=1,或,a,=4.,d,=,-,5,12,a,n,=1,或,a,n,=,-,n,+,.,5,12,5,32,经验证知,a,n,=1,时,S,n,=5;,另一个情况时,S,n,=,-,4,均合题意,.,a,n,=1,或,a,n,=,-,n,+,即为所求数列,a,n,通项公式,.,5,12,5,32,10/18,解法,2:,S,n,是等差数列前,n,项和,故可设,S,n,=,an,2,+,bn,依题意得,:,1,3,1,4,(,a,3,2,+,b,3),(,a,4,2,+,b,4)=(,a,5,2,+,b,5),2,1,4,(,a,3,2,+,b,3),+,(,a,4,2,+,b,4)=2.,1,3,25,1,解得,a,=0,b,=1,或,b,=.,a,=,-,5,26,6,5,S,n,=,n,或,S,n,=,-,n,2,+,n,.,5,26,6,5,在等差数列中,n,2,时,a,n,=,S,n,-,S,n,-,1,a,1,亦适合公式,.,a,n,=1,或,a,n,=,-,n,+,.,5,12,5,32,整理得,13,a,2,+3,ab,=0,7,a,+2,b,=2.,5.,设,S,n,是等差数列,a,n,前,n,项和,.,已知,S,3,与,S,4,等比中项为,S,5,S,3,与,S,4,等差中项为,1,求等差数列,a,n,通项,a,n,.,1,5,1,3,1,4,1,3,1,4,11/18,解法,3:,S,n,是等差数列前,n,项和,数列,是等差数列,.,S,n,n,+=2,S,3,3,S,5,5,S,4,4,依题意得,:,=(),2,S,5,5,S,3,3,S,4,4,+=2,S,3,3,S,4,4,解得,:,S,4,=4,S,3,=3,S,5,=5,或,S,4,=,S,3,=,S,5,=,-,4,8,5,5,24,a,4,=,S,4,-,S,3,=1,a,5,=,S,5,-,S,4,=1,或,a,4,=,-,a,5,=,-,.,5,28,5,16,a,n,=1,或,a,n,=,-,n,+,.,5,12,5,32,5.,设,S,n,是等差数列,a,n,前,n,项和,.,已知,S,3,与,S,4,等比中项为,S,5,S,3,与,S,4,等差中项为,1,求等差数列,a,n,通项,a,n,.,1,5,1,3,1,4,1,3,1,4,12/18,解法,4:,依题意,S,3,=3,a,2,S,4,=2(,a,2,+,a,3,),S,5,=5,a,3,整理得,:,3,a,2,+,a,3,=4,a,2,(,a,2,+,a,3,)=2,a,3,2,代入,S,5,5,S,3,3,S,4,4,+=2,S,3,3,S,4,4,=(),2,4,5,解得,a,2,=1,a,3,=1,或,a,3,=,-,.,a,2,=,8,5,a,n,=1,或,a,n,=,-,n,+,.,5,12,5,32,5.,设,S,n,是等差数列,a,n,前,n,项和,.,已知,S,3,与,S,4,等比中项为,S,5,S,3,与,S,4,等差中项为,1,求等差数列,a,n,通项,a,n,.,1,5,1,3,1,4,1,3,1,4,13/18,6.,已知,a,n,+1,=2+,a,n,(,n,N,+,),且,a,1,=,a,求,a,n,.,1,2,解,:,a,1,=,a,a,2,=2+,a,1,2,=4,-,2,1,+2,-,1,a,故猜测,:,a,n,=4,-,2,3,-,n,+2,1,-,n,a,用数学归纳法证实以下,:,a,5,=2+,a,4,1,2,a,3,=2+,a,2,=3+,a,1,2,1,4,=4,-,2,0,+2,-,2,a,a,4,=2+,a,3,=+,a,1,2,7,2,1,8,=4,-,2,-,1,+2,-,3,a,=4,-,2,-,2,+2,-,4,a,=4,-,2,2,+2,0,a,证实从略,.,故,a,n,=4,-,2,3,-,n,+2,1,-,n,a,.,解法二,:,结构等比数列求解,(,略,),.,14/18,7.,设,数列,a,n,是公差不为,0,等差数列,S,n,是数列,a,n,前,n,项和,且,S,3,2,=9,S,2,S,4,=4,S,2,求数列,a,n,通项公式,.,解,:,设等差数列,a,n,公差为,d,由,S,n,=,na,1,+,及已知条件得,:,n,(,n,-,1),d,2,(3,a,1,+3,d,),2,=9,(2,a,1,+,d,),4,a,1,+6,d,=4,(2,a,1,+,d,),由,得,:,d,=2,a,1,代入,有,:9,a,1,2,=4,a,1,.,解得,:,a,1,=0,或,a,1,=.,4,9,当,a,1,=0,时,d,=0,与已知条件矛盾,舍去,;,当,a,1,=,时,d,=.,4,9,8,9,a,n,=+(,n,-,1)=,n,-,.,4,9,8,9,4,9,8,9,故数列,a,n,通项公式为,a,n,=,n,-,.,4,9,8,9,15/18,8.,已知数列,a,n,是等差数列,且,a,1,=2,a,1,+,a,2,+,a,3,=12,(1),求数列,a,n,通项公式,;(2),令,b,n,=,a,n,3,n,求数列,b,n,前,n,项和公式,.,解,:,(1),设数列,a,n,公差为,d,则由已知得,3,a,1,+3,d,=12,d,=2.,a,n,=2+(,n,-,1),2=2,n,.,故数列,a,n,通项公式,为,a,n,=2,n,.,(2),由,b,n,=,a,n,3,n,=2,n,3,n,得数列,b,n,前,n,项和,S,n,=2,3+4,3,2,+,+(2,n,-,2),3,n,-,1,+2,n,3,n,3,S,n,=2,3,2,+4,3,3,+,+(2,n,-,2),3,n,+2,n,3,n,+1,将,式减,式得,:,-,2,S,n,=2(3+3,2,+,+3,n,),-,2,n,3,n,+1,=3(3,n,-,1),-,2,n,3,n,+1,.,S,n,=+,n,3,n,+1,.,3(1,-,3,n,),2,又,a,1,=2,16/18,再见,17/18,邢台月嫂,邢台月嫂,frsxfxen,再见！,18/18,