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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,第三級,第四級,第五級,*,等差數(shù)列與差比數(shù)列的通項公式,類型一:等差數(shù)列與等比數(shù)列的通項:,公式,練習(xí):,類型二:類等差(比)數(shù)列,方法:,累加(乘),一、若數(shù)列有形如,a,n,1,a,n,f,(,n,)的解析式,而,f,(1),f,(2),f,(,n,)的和是可求的,則可用多式累(迭)加法求得,a,n,.,(2011年廈門質(zhì)檢),已知數(shù)列,a,n,中,,a,1,20,,a,n,1,a,n,2,n,1,,n,N,*,,則數(shù)列,a,n,的通項公式,a,n,_.,解析,:,由條件,a,n,1,a,n,2,n,1,,n,N,*,,,即,a
2、,n,1,a,n,2,n,1,得,a,2,a,1,1,,a,3,a,2,3,,a,4,a,3,5,,,,a,n,1,a,n,2,2,n,5,,a,n,a,n,1,2,n,3,,以上,n,1個式子相加并化簡,得,a,n,a,1,(,n,1),2,n,2,2,n,21.,答案,:,n,2,2,n,21,變式探究,1已知數(shù)列,a,n,中,,a,1,1,,a,n,1,a,n,2,n,,求,a,n,.,解析,:,當(dāng),n,2時,,a,2,a,1,2,,a,3,a,2,2,2,,,a,4,a,3,2,3,,,,,a,n,a,n,1,2,n,1,.,將這,n,1個式子累加起來可得,a,n,a,1,22,2,2
3、,n,1,,,a,n,a,1,22,2,2,n,1,122,2,2,n,1,2,n,1.,當(dāng),n,1時,,a,1,適合上式,故,a,n,2,n,1.,二、若數(shù)列有形如,a,n,f,(,n,),a,n,1,的解析關(guān)系,而,f,(1),f,(2),f,(,n,)的積是可求的,則可用多式累(迭)乘法求得,a,n,.,設(shè),a,n,的首項為1的正項數(shù)列,且,n,a,n,1,a,n,0,求它的通項公式,解析,:,由題意,a,1,1 ,a,n,0,(,n,1,2,3,,),,方法二:,練習(xí),由整理得,再用累乘法 也可以,練習(xí),類型五:待定系數(shù)法求數(shù)列的通項:,則可考慮待定系數(shù)法設(shè),構(gòu)造新的輔助數(shù)列,是首項為
4、,公比為,q,的等比數(shù)列,求出,再進一步求通項,若數(shù)列有形如,a,n,pa,n,1,q,(,n,2,,p,,,q,為常數(shù),,pq,0,,p,1)的線性遞推關(guān)系,則可用待定系數(shù)法求得,a,n,.,具體思路,:設(shè)遞推式可化為,a,n,1,A,p,(,a,n,A,),,得,a,n,1,pa,n,(,p,1),A,,與已知遞推式比較,解得,A, ,故可將遞推式化為,a,n, ,p,(,a,n-1,+,),,,構(gòu)造數(shù)列,b,n,,其中,b,n,a,n, , 則,b,n,1,pb,n,,即 ,p,,所以,b,n,為等比數(shù)列故可求出,b,n,f,(,n,),再將,b,n,a,n, 代入即可得,a,n,.,已
5、知數(shù)列,a,n,中,,a,1,1,,a,n,1,a,n,1,求,a,n,.,解析,:,解法一,:,數(shù)列,b,n,為等比數(shù)列,又,a,1,32,,點評,:(1)注意數(shù)列解題中的換元思想的運用,如,b,n,a,n,3.,(2)對數(shù)列遞推式,a,n,1,pa,n,q,,我們通常將其化為,p,,設(shè),b,n,a,n,A,,構(gòu)造數(shù)列,b,n,為等比數(shù)列,,,練習(xí),四、遞推式如,a,n,pa,n,1,rq,n,(,n,2,,pqr,0,,p,,,q,,,r,為常數(shù))型的通項的求法,具體思路,:1.等式兩邊同除以,q,n,,,已知數(shù)列,a,n,滿足,a,n,4,a,n,1,2,n,(,n,2,,n,N,*,)
6、,且,a,1,2.求,a,n,.,解析,:,解法一,: ,a,n,4,a,n,1,2,n,解法二,:,a,n,4,a,n,1,2,n,,,令,a,n,2,n,4(,a,n,1,2,n,1,),(,n,2),,得,a,n,4,a,n,1,2,n,,與已知遞推式比較得,1,,a,n,2,n,4 ,,又,a,1,2,21,4,,a,n,2,n,是首項為4,公比為4的等比數(shù)列,a,n,2,n,44,n,1,,,a,n,4,n,2,n,2,2,n,2,n,.,練習(xí),變式探究,5,(2011年鹽城模擬),在數(shù)列,a,n,中,,a,1,2,,a,n,1,a,n,n,1,(2,)2,n,(,n,N,*,),其
7、中,0.求數(shù)列,a,n,的通項公式,解析,:,由,a,n,1,a,n,n,1,(2,)2,n,(,n,N,*,),,0,,得,a,n,1,a,n,n,1,2,n,1,2,n,,,所以數(shù)列,a,n,的通項公式為,a,n,(,n,1),n,2,n,.,方法二,:,累加,由得,五、遞推式如,a,n,pa,n,1,qn,r,(,n,2,,pq,0,,p,,,q,為常數(shù))型數(shù)列的通項求法,具體思路,:等價轉(zhuǎn)化為,a,n,xn,y,p,(,a,n,1,x,(,n,1),y,),再化為,a,n,pa,n,1,(,p,1),xn,(,p,1),y,,比較對應(yīng)系數(shù),解出,x,,,y,,進而轉(zhuǎn)化為例3的數(shù)列,(2
8、011,年濟寧模擬,)已知數(shù)列,a,n,中,,a,1, ,點(,n,2,a,n,1,a,n,)在直線,y,x,上,其中,n,1,2,3,.求數(shù)列,a,n,的通項,解析,:,點(,n,2,a,n,1,a,n,)在直線,y,x,上,,2,a,n,1,a,n,n,.,令,a,n,1,x,(,n,1),y, (,a,n,nx,y,),可化為,2,a,n,1,a,n,xn,2,x,y,0與,比較系數(shù)得,x,1,,y,2., 可化為,a,n,1,(,n,1)2 (,a,n,n,2),,變式探究,6(2010,年豐臺區(qū)模擬,)在數(shù)列,a,n,中,,a,1,2,,a,n,1,4,a,n,3,n,1,,n,N,
9、*,.,(1)設(shè),b,n,a,n,n,,求數(shù)列 的通項;,(2)求數(shù)列,a,n,的前,n,項和,S,n,.,解析,:,(1)由題設(shè),a,n,1,4,a,n,3,n,1,,得,a,n,1,(,n,1)4(,a,n,n,),,n,N,*,.,b,n,a,n,n,,,b,n,1,a,n,1,(,n,1),,b,n,1,4,b,n,.,又,b,1,a,1,11,所以數(shù)列 是首項為1,且公比為4的等比數(shù)列,b,n,4,n,1,.,(2)由(1)可知,a,n,n,4,n,1,,于是數(shù)列,a,n,的通項公式為,a,n,4,n,1,n,.,七、倒數(shù)法求通項,(1)對于遞推式如,a,n,1,pa,n,qa,n,
10、1,a,n,(,p,,,q,為常數(shù),,pq,0)型的數(shù)列,求其通項公式,具體思路,:兩端除以,a,n,1,a,n,得: ,p,q,,,若,p,1,則構(gòu)成以首項為 ,公差為,q,的等差數(shù)列 ;,若,p,1,轉(zhuǎn)化為例3求解,(2011年保定摸底),已知數(shù)列,a,n,滿足,a,1,1,,n,2時,,a,n,1,a,n,2,a,n,1,a,n,,求通項公式,a,n,.,解析,:,a,n,1,a,n,2,a,n,1,a,n,,,變式探究,答案,:,a,n,(2)若數(shù)列,a,n,有形如,a,n,1, 的關(guān)系,求其通項的具體思路是:取倒數(shù)后得 ,即化為例3的數(shù)列,求出 ,再求得,a,n,.,設(shè)數(shù)列,a,n,
11、滿足,a,1,2,,a,n,1, (,n,N,*,),,求,a,n,.,解析,:,由,a,n,1,取倒數(shù),,類型六:特征根法求數(shù)列通。,(條件:若,的,相鄰兩項關(guān)系式可化為,可用這種方法;(其中方程,該數(shù)列的特征根),的根稱為,(一)有,兩特根,與,,可令,構(gòu)造等比數(shù)列,,則可,進而求出,等比數(shù)列通項公式求出,特征根為0與1,略解:依題意可得該數(shù)列特征根為0與1,練習(xí),(改編),構(gòu)造輔助數(shù)列 ,分析,(二)有,一根,時,可令,易得 是,等差,數(shù)列,求,進而求出,唯一特征根1,解:依題意可得該數(shù)列有惟一特征根為1,該題也可以先求出前幾項,,再猜想歸納出其通項,但要特別注意要用數(shù)學(xué)歸納法證明。,練
12、習(xí),(三)沒有特征根,則可由遞推關(guān)系式得出若干項可判斷,是周期數(shù)列,(題型)若數(shù)列相鄰三項的關(guān)系可化為,且方程,有解,則可用待定系數(shù)法設(shè),公比的輔助等比數(shù)列,構(gòu)造新的以y為,,轉(zhuǎn)化相鄰兩項處理;,若,有兩組值,也可得到兩個等比數(shù)列,分別求其通項,再由方程組求出,兩種情況一起考慮,即,累加,方程思想,分別得到:,由得,練習(xí),【解析】(1)由求根公式,不妨設(shè),(2),(3),遞推式如,的數(shù)列通項的求法,【,具體思路,】,若p=1,則等式兩邊取常用對數(shù)或自然對數(shù),化為: ,得到首項為 ,公比為r的等比數(shù)列 ,所以 = ,得,若p1,則等式兩邊取以p為底的對數(shù)得:,轉(zhuǎn)化為題型三求通項。,(,11年石家莊市模擬,)若數(shù)列a,n,中,,且 ,則數(shù)列的通項公式,為_,【解析】 及 知,兩邊取常用對數(shù)得:, 是以首項為 ,公比為2的,等比數(shù)列。,其他方法:有構(gòu)造常數(shù)數(shù)列,取對數(shù)(注意真數(shù)大于零),取倒數(shù),歸納法(注意要用數(shù)學(xué)歸納法證明),左邊能否因式分式?,累乘法,特征根法,