,高考大题,增分专项,高考大题增分专项三高考中的数列,考情分析,典例突破,专题总结,高考大题增分专项三,高考,中的数列,高考大题增分专项三高考中的数列,-,2,-,从近五年高考试题分析来看,高考数列解答题主要题型有,:,等差、等比数列的综合问题,;,证明一个数列为等差或等比数列,;,求数列的通项及非等差、等比数列的前,n,项和,;,证明数列型不等式,.,命题特点是试题题型规范、方法可循、难度稳定在中档,.,-2-从近五年高考试题分析来看,高考数列解答题主要题型有:等,-,3,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,突破策略一,公式法,对于等差、等比数列,求其通项及求前,n,项的和时,只需利用等差数列或等比数列的通项公式及求和公式求解即可,.,-3-题型一题型二题型三题型四题型五策略一策略二突破策略一,-,4,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,(1),求数列,a,n,的通项公式,;,(2),求数列,b,n,的前,n,项和,S,n,.,a,1,=,4,a,n,是首项为,4,公差为,3,的等差数列,.,a,n,=,4,+,(,n-,1)3,=,3,n+,1,.,(2),由,(1),及,a,n,b,n+,1,=nb,n,+b,n+,1,例,1,已知,a,n,是公差为,3,的等差数列,数列,b,n,满足,-4-题型一题型二题型三题型四题型五策略一策略二(1)求数列,-,5,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,对点训练,1,在等比数列,a,n,中,已知,a,1,=,2,a,4,=,16,.,(1),求数列,a,n,的通项公式,.,(2),若,a,3,a,5,分别为等差数列,b,n,的第,4,项和第,16,项,试求数列,b,n,的通项公式及前,n,项和,S,n,.,-5-题型一题型二题型三题型四题型五策略一策略二对点训练1在,-,6,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,突破策略二,转化法,无论是求数列的通项还是求数列的前,n,项和,通过变形、整理后,能够把数列转化为等差数列或等比数列,进而利用等差数列或等比数列的通项公式或求和公式解决问题,.,-6-题型一题型二题型三题型四题型五策略一策略二突破策略二,-,7,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,例,2,已知等比数列,a,n,的前,n,项和为,S,n,a,1,=,3,且,3,S,1,2,S,2,S,3,成等差数列,.,(1),求数列,a,n,的通项公式,;,(2),设,b,n,=,log,3,a,n,T,2,n,=b,1,b,2,-b,2,b,3,+b,3,b,4,-b,4,b,5,+,+b,2,n-,1,b,2,n,-b,2,n,b,2,n+,1,求,T,2,n,.,解,:,(,1),3,S,1,2,S,2,S,3,成等差数列,4,S,2,=,3,S,1,+S,3,.,4(,a,1,+a,2,),=,3,a,1,+,(,a,1,+a,2,+a,3,),即,a,3,=,3,a,2,.,公比,q=,3,.,a,n,=a,1,q,n-,1,=,3,n,.,(2),由,(1),知,b,n,=,log,3,a,n,=,log,3,3,n,=n,b,2,n-,1,b,2,n,-b,2,n,b,2,n+,1,=,(2,n-,1)2,n-,2,n,(2,n+,1),=-,4,n,T,2,n,=,(,b,1,b,2,-b,2,b,3,),+,(,b,3,b,4,-b,4,b,5,),+,+,(,b,2,n-,1,b,2,n,-b,2,n,b,2,n+,1,),-7-题型一题型二题型三题型四题型五策略一策略二例2已知等比,-,8,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,对点训练,2,设,a,n,是公比大于,1,的等比数列,S,n,为数列,a,n,的前,n,项和,已知,S,3,=,7,且,a,1,+,3,3,a,2,a,3,+,4,成等差数列,.,(1),求数列,a,n,的通项公式,;,(2),令,b,n,=,ln,a,3,n+,1,n=,1,2,求数列,b,n,的前,n,项和,T,n,.,-8-题型一题型二题型三题型四题型五策略一策略二对点训练2设,-,9,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,-9-题型一题型二题型三题型四题型五策略一策略二,-,10,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,突破策略一,定义法,用,定义法证明一个数列是等差数列,常采用的两个式子,a,n,-a,n-,1,=,d,(,n,2),和,a,n+,1,-a,n,=d,前者必须加上,“,n,2”,否则,n=,1,时,a,0,无意义,;,用定义法证明一个数列是等比数列也常采用两个式子,-10-题型一题型二题型三题型四题型五策略一策略二突破策略一,-,11,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,(1),求,a,1,a,2,;,(2),求数列,a,n,的通项公式,并证明数列,a,n,是等差数列,;,(3),若数列,b,n,满足,a,n,=,log,2,b,n,试证明数列,b,n,是等比数列,并求其前,n,项和,T,n,.,又,a,1,=,5,满足,a,n,=,3,n+,2,故,a,n,=,3,n+,2,.,因为,a,n+,1,-a,n,=,3(,n+,1),+,2,-,(3,n+,2),=,3(,n,N,*,),所以数列,a,n,是以,5,为首项,3,为公差的等差数列,.,-11-题型一题型二题型三题型四题型五策略一策略二(1)求a,-,12,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,-12-题型一题型二题型三题型四题型五策略一策略二,-,13,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,对点训练,3,已知,数列,a,n,其前,n,项和为,S,n,满足,a,1,=,2,S,n,=,na,n,+,a,n-,1,其中,n,2,n,N,*,R,.,(1),若,=,0,=,4,b,n,=a,n+,1,-,2,a,n,(,n,N,*,),求证,:,数列,b,n,是等比数列,;,证明,:,(1),若,=,0,=,4,则,S,n,=,4,a,n-,1,(,n,2),所以,a,n+,1,=S,n+,1,-S,n,=,4(,a,n,-a,n-,1,),即,a,n+,1,-,2,a,n,=,2(,a,n,-,2,a,n-,1,),所以,b,n,=,2,b,n-,1,.,又由,a,1,=,2,a,1,+a,2,=,4,a,1,得,a,2,=,3,a,1,=,6,a,2,-,2,a,1,=,20,即,b,n,0,-13-题型一题型二题型三题型四题型五策略一策略二对点训练3,-,14,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,(2),若,a,2,=,3,由,a,1,+a,2,=,2,a,2,+,a,1,得,5,=,6,+,2,.,即,(,n-,1),a,n+,1,-,(,n-,2),a,n,-,2,a,n-,1,=,0,所以,na,n+,2,-,(,n-,1),a,n+,1,-,2,a,n,=,0,两式相减得,na,n+,2,-,2(,n-,1),a,n+,1,+,(,n-,2),a,n,-,2,a,n,+,2,a,n-,1,=,0,所以,n,(,a,n+,2,-,2,a,n+,1,+a,n,),+,2(,a,n+,1,-,2,a,n,+a,n-,1,),=,0,-14-题型一题型二题型三题型四题型五策略一策略二(2)若a,-,15,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,因为,a,1,-,2,a,2,+a,3,=,0,所以,a,n+,2,-,2,a,n+,1,+a,n,=,0,即数列,a,n,是等差数列,.,-15-题型一题型二题型三题型四题型五策略一策略二因为a1-,-,16,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,突破策略二,递推相减化归法,对已知数列,a,n,与,S,n,的关系,证明,a,n,为等差或等比数列的问题,解题思路为,:,由,a,n,与,S,n,的关系递推出,n,为,n+,1,时的关系式,两关系式相减后,进行化简、整理,最终化归为用定义法证明,.,-16-题型一题型二题型三题型四题型五策略一策略二突破策略二,-,17,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,例,4,已知数列,a,n,的前,n,项和为,S,n,S,n,=,(,m+,1),-ma,n,对任意的,n,N,*,都成立,其中,m,为常数,且,m-,1,.,(1),求证,:,数列,a,n,是等比数列,;,(2),记数列,a,n,的公比为,q,设,q=f,(,m,),若数列,b,n,满足,(,3),在,(2),的条件下,设,c,n,=b,n,b,n+,1,数列,c,n,的前,n,项和为,T,n,求证,:,T,n,1,.,-17-题型一题型二题型三题型四题型五策略一策略二例4已知数,-,18,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,证明,:,(,1),当,n=,1,时,a,1,=S,1,=,1,.,S,n,=,(,m+,1),-ma,n,S,n-,1,=,(,m+,1),-ma,n-,1,(,n,2),由,-,得,a,n,=ma,n-,1,-ma,n,(,n,2),即,(,m+,1),a,n,=ma,n-,1,.,a,1,0,m,0,解得,q=,2,.,所以,b,n,=,2,n,.,由,b,3,=a,4,-,2,a,1,可得,3,d-a,1,=,8,.,由,S,11,=,11,b,4,可得,a,1,+,5,d=,16,联立,解得,a,1,=,1,d=,3,由此可得,a,n,=,3,n-,2,.,所以数列,a,n,的通项公式为,a,n,=,3,n-,2,数列,b,n,的通项公式为,b,n,=,2,n,.,-26-题型一题型二题型三题型四题型五策略一策略二对点训练5,-,27,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,(2),设数列,a,2,n,b,2,n-,1,的前,n,项和为,T,n,由,a,2,n,=,6,n-,2,b,2,n-,1,=,24,n-,1,有,a,2,n,b,2,n-,1,=,(3,n-,1)4,n,故,T,n,=,24+54,2,+84,3,+,+,(3,n-,1)4,n,4,T,n,=,24,2,+54,3,+,84,4,+,+,(3,n-,4)4,n,+,(3,n-,1)4,n+,1,上述两式相减,得,-,3,T,n,=,24+34,2,+34,3,+,+,34,n,-,(3,n-,1)4,n+,1,-27-题型一题型二题型三题型四题型五策略一策略二(2)设数,-,28,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,突破策略二,裂项相消法,把数列的通项拆成两项之差,在求和时中间的一些项可以相互抵消,从而求得其和,.,利用裂项相消法求和时,要注意抵消后所剩余的项是前后对称的,.,-28-题型一题型二题型三题型四题型五策略一策略二突破策略二,-,29,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,-29-题型一题型二题型三题型四题型五策略一策略二,-,30,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,-30-题型一题型二题型三题型四题型五策略一策略二,-,31,-,题型一,题型二,题型三,题型四,题型五,策略一,策略二,对点训练,6,(2019,河北唐山高三第一次模拟考试,),已知数列,a,n,的前,n,项和为,S,n,且,(1),求,S,n,a,n,;,当,n,2,时,a,n,=S,n,-S,n-,1,=,2,n-,1,当,n=,1,时,a,1,=,1,适合上式,所以,a,n,=,2,n-,1,.,-31-题型一题型二题型三题型四题型五策略一策略二对点训练6,-,32,-,