《高考數(shù)學(xué)（福建專用）一輪復(fù)習(xí)課件：高考大題增分專項(xiàng)三 高考中的數(shù)列(共43張PPT)》由會(huì)員分享，可在線閱讀，更多相關(guān)《高考數(shù)學(xué)（福建專用）一輪復(fù)習(xí)課件：高考大題增分專項(xiàng)三 高考中的數(shù)列(共43張PPT)（43頁珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、單擊此處編輯母版標(biāo)題樣式,單擊此處編輯母版文本樣式,第二級(jí),第三級(jí),第四級(jí),第五級(jí),2019/8/19,#,高考大題增分專項(xiàng)三,高考,中的數(shù)列,-,2,-,從近五年高考試題分析來看,高考數(shù)列解答題主要題型有,:,等差、等比數(shù)列的綜合問題,;,證明一個(gè)數(shù)列為等差或等比數(shù)列,;,求數(shù)列的通項(xiàng)及非等差、等比數(shù)列的前,n,項(xiàng)和,;,證明數(shù)列型不等式,.,命題特點(diǎn)是試題題型規(guī)范、方法可循、難度穩(wěn)定在中檔,.,-,3,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,突破策略一,公式法,對(duì)于等差、等比數(shù)列,求其通項(xiàng)及求前,n,項(xiàng)的和時(shí),只需利用等差數(shù)列或等比數(shù)列的通項(xiàng)公式及求和公式求解即可,.,
2、-,4,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,例,1,(2018,山東淄博一模,),已知,a,n,是公差為,3,的等差數(shù)列,數(shù)列,b,n,(1),求數(shù)列,a,n,的通項(xiàng)公式,;,(2),求數(shù)列,b,n,的前,n,項(xiàng)和,S,n,.,a,1,=,4,a,n,是首項(xiàng)為,4,公差為,3,的等差數(shù)列,.,a,n,=,4,+,(,n-,1),3,=,3,n+,1,.,(2),由,(1),及,a,n,b,n+,1,=nb,n,+b,n+,1,-,5,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,對(duì)點(diǎn)訓(xùn)練,1,在等比數(shù)列,a,n,中,已知,a,1,=,2,a,4,=,16,
3、.,(1),求數(shù)列,a,n,的通項(xiàng)公式,.,(2),若,a,3,a,5,分別為等差數(shù)列,b,n,的第,4,項(xiàng)和第,16,項(xiàng),試求數(shù)列,b,n,的通項(xiàng)公式及前,n,項(xiàng)和,S,n,.,-,6,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,突破策略二,轉(zhuǎn)化法,無論是求數(shù)列的通項(xiàng)還是求數(shù)列的前,n,項(xiàng)和,通過變形、整理后,能夠把數(shù)列轉(zhuǎn)化為等差數(shù)列或等比數(shù)列,進(jìn)而利用等差數(shù)列或等比數(shù)列的通項(xiàng)公式或求和公式解決問題,.,-,7,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,例,2,已知等比數(shù)列,a,n,的前,n,項(xiàng)和為,S,n,a,1,=,3,且,3,S,1,2,S,2,S,3
4、,成等差數(shù)列,.,(1),求數(shù)列,a,n,的通項(xiàng)公式,;,(2),設(shè),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,成等差數(shù)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,=,l
5、og,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,),-,8,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,對(duì)點(diǎn)訓(xùn)練,2,設(shè),a,n,是公比大于,1,的等比數(shù)列,S,n,為數(shù)列,a,n,的前,n,項(xiàng)和,已知,S,3,=,7,且,a,1,+,3,3,a,2,a,3,+,4,成等差
6、數(shù)列,.,(1),求數(shù)列,a,n,的通項(xiàng)公式,;,(2),令,b,n,=,ln,a,3,n+,1,n=,1,2,求數(shù)列,b,n,的前,n,項(xiàng)和,T,n,.,-,9,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,-,10,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,突破策略一,定義法,用,定義法證明一個(gè)數(shù)列是等差數(shù)列,常采用的兩個(gè)式子,a,n,-a,n-,1,=d,(,n,2),和,a,n+,1,-a,n,=d,前者必須加上,“,n,2”,否則,n=,1,時(shí),a,0,無意義,;,用定義法證明一個(gè)數(shù)列是等比數(shù)列也常采用兩個(gè)式子,-,11,-,題型一,題型二,題型三,題
7、型四,題型五,策略一,策略二,(1),求,a,1,a,2,;,(2),求數(shù)列,a,n,的通項(xiàng)公式,并證明數(shù)列,a,n,是等差數(shù)列,;,(3),若數(shù)列,b,n,滿足,a,n,=,log,2,b,n,試證明數(shù)列,b,n,是等比數(shù)列,并求其前,n,項(xiàng)和,T,n,.,又,a,1,=,5,滿足,a,n,=,3,n+,2,故,a,n,=,3,n+,2,.,因?yàn)?a,n+,1,-a,n,=,3(,n+,1),+,2,-,(3,n+,2),=,3(,n,N,*,),所以數(shù)列,a,n,是以,5,為首項(xiàng),3,為公差的等差數(shù)列,.,-,12,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,-,13,-,
8、題型一,題型二,題型三,題型四,題型五,策略一,策略二,對(duì)點(diǎn)訓(xùn)練,3,(2018,江蘇淮安一模節(jié)選,),已知數(shù)列,a,n,其前,n,項(xiàng)和為,S,n,滿足,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,*,),求證,:,數(shù)列,b,n,是等比數(shù)列,;,證明,:,(1),若,=,0,=,4,則,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
9、,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,-,14,-,題型一,題型二,題型三,題型四,題型五,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,=,0,兩式相減得,na,n+,2,-,2(,n-,1),
10、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,-,15,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,因?yàn)?a,1,-,2,a,2,+a,3,=,0,所以,a,n+,2,-,2,a,n+,1,+a,n,=,0,即數(shù)列,a,n,是等差數(shù)列,.,-,16,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,突破策略二,遞推相減化歸法,對(duì)已知數(shù)列,a,n,與,S,n,的關(guān)系,證明,a,n,為等差或等比數(shù)列的
11、問題,解題思路為,:,由,a,n,與,S,n,的關(guān)系遞推出,n,為,n+,1,時(shí)的關(guān)系式,兩關(guān)系式相減后,進(jìn)行化簡(jiǎn)、整理,最終化歸為用定義法證明,.,-,17,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,例,4,已知數(shù)列,a,n,的前,n,項(xiàng)和為,S,n,S,n,=,(,m+,1),-ma,n,對(duì)任意的,n,N,*,都成立,其中,m,為常數(shù),且,m-,1,.,(1),求證,:,數(shù)列,a,n,是等比數(shù)列,;,(2),記數(shù)列,a,n,的公比為,q,設(shè),q=f,(,m,),若數(shù)列,b,n,滿足,(,3),在,(2),的條件下,設(shè),c,n,=b,n,b,n+,1,數(shù)列,c,n,的前,n
12、,項(xiàng)和為,T,n,求證,:,T,n,1,.,-,18,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,證明,(1),當(dāng),n=,1,時(shí),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=,1
13、6,聯(lián)立,解得,a,1,=,1,d=,3,由此可得,a,n,=,3,n-,2,.,所以數(shù)列,a,n,的通項(xiàng)公式為,a,n,=,3,n-,2,數(shù)列,b,n,的通項(xiàng)公式為,b,n,=,2,n,.,-,26,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,(2),設(shè)數(shù)列,a,2,n,b,2,n-,1,的前,n,項(xiàng)和為,T,n,由,a,2,n,=,6,n-,2,b,2,n-,1,=,2,4,n-,1,有,a,2,n,b,2,n-,1,=,(3,n-,1),4,n,故,T,n,=,2,4,+,5,4,2,+,8,4,3,+,+,(3,n-,1),4,n,4,T,n,=,2,4,2,+,5,4
14、,3,+,8,4,4,+,+,(3,n-,4),4,n,+,(3,n-,1),4,n+,1,上述兩式相減,得,-,3,T,n,=,2,4,+,3,4,2,+,3,4,3,+,+,3,4,n,-,(3,n-,1),4,n+,1,-,27,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,突破策略二,裂項(xiàng)相消法,把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差,在求和時(shí)中間的一些項(xiàng)可以相互抵消,從而求得其和,.,利用裂項(xiàng)相消法求和時(shí),要注意抵消后所剩余的項(xiàng)是前后對(duì)稱的,.,-,28,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,-,29,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,
15、-,30,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,對(duì)點(diǎn)訓(xùn)練,6,已知,數(shù)列,a,n,的前,n,項(xiàng)和為,S,n,a,1,=,1,且,3,S,n,=a,n+,1,-,1,.,(1),求數(shù)列,a,n,的通項(xiàng)公式,;,解,:,(1),3,S,n,=a,n+,1,-,1,當(dāng),n,1,時(shí),3,S,n-,1,=a,n,-,1,-,得,3(,S,n,-S,n-,1,),=,3,a,n,=a,n+,1,-a,n,則,a,n+,1,=,4,a,n,又,a,2,=,3,a,1,+,1,=,4,=,4,a,1,數(shù)列,a,n,是首項(xiàng)為,1,公比為,4,的等比數(shù)列,則,a,n,=,4,n-,1,.,-
16、,31,-,題型一,題型二,題型三,題型四,題型五,策略一,策略二,(2),由,(1),得,a,2,=,4,S,3,=,21,-,32,-,題型一,題型二,題型三,題型四,題型五,要,證明關(guān)于一個(gè)數(shù)列的前,n,項(xiàng)和的不等式,一般有兩種思路,:,一是先求和再對(duì)和式放縮,;,二是先對(duì)數(shù)列的通項(xiàng)放縮再求數(shù)列的和,必要時(shí)對(duì)其和再放縮,.,-,33,-,題型一,題型二,題型三,題型四,題型五,-,34,-,題型一,題型二,題型三,題型四,題型五,對(duì)點(diǎn)訓(xùn)練,7,(2018,天津部分區(qū)質(zhì)量調(diào)查,),已知數(shù)列,a,n,為等比數(shù)列,數(shù)列,b,n,為等差數(shù)列,且,b,1,=a,1,=,1,b,2,=a,1,+a,2,a,3,=,2,b,3,-,6,.,(1),求數(shù)列,a,n,b,n,的通項(xiàng)公式,;,(1),解,:,設(shè)數(shù)列,a,n,的公比為,q,數(shù)列,b,n,的公差為,d,所以,a,n,=,2,n-,1,b,n,=,2,n-,1,.,-,35,-,題型一,題型二,題型三,題型四,題型五,-,36,-,題型一,題型二,題型三,題型四,題型五,求解,數(shù)列中的存在性問題,先假設(shè)所探求對(duì)象存在,再以此假設(shè)為前提條件
高考數(shù)學(xué)（福建專用）一輪復(fù)習(xí)課件：高考大題增分專項(xiàng)三 高考中的數(shù)列(共43張PPT)
