《高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 4.2.1 等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和課件 文》由會(huì)員分享，可在線閱讀，更多相關(guān)《高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 4.2.1 等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和課件 文（23頁珍藏版）》請(qǐng)?jiān)谘b配圖網(wǎng)上搜索。
1、4 4. .2 2. .1 1等差、等比數(shù)列與數(shù)等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和列的通項(xiàng)及求和-2-等差、等比數(shù)列的通項(xiàng)及求和等差、等比數(shù)列的通項(xiàng)及求和例1(2018全國,文17)記Sn為等差數(shù)列an的前n項(xiàng)和,已知a1=-7,S3=-15.(1)求an的通項(xiàng)公式;(2)求Sn,并求Sn的最小值.解 (1)設(shè)an的公差為d,由題意得3a1+3d=-15.由a1=-7得d=2.所以an的通項(xiàng)公式為an=2n-9.(2)由(1)得Sn=n2-8n=(n-4)2-16.所以當(dāng)n=4時(shí),Sn取得最小值,最小值為-16.解題心得對(duì)于等差、等比數(shù)列,求其通項(xiàng)及前n項(xiàng)和時(shí),只需利用等差數(shù)列或等比數(shù)列的通項(xiàng)公
2、式及求和公式求解即可.-3-對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練1已知等差數(shù)列an的前n項(xiàng)和為Sn,等比數(shù)列bn的前n項(xiàng)和為Tn,a1=-1,b1=1,a2+b2=2.(1)若a3+b3=5,求bn的通項(xiàng)公式;(2)若T3=21,求S3.解 設(shè)an的公差為d,bn的公比為q,則an=-1+(n-1)d,bn=qn-1.由a2+b2=2得d+q=3.(1)由a3+b3=5,得2d+q2=6.因此bn的通項(xiàng)公式為bn=2n-1.(2)由b1=1,T3=21得q2+q-20=0,解得q=-5或q=4.當(dāng)q=-5時(shí),由得d=8,則S3=21.當(dāng)q=4時(shí),由得d=-1,則S3=-6.-4-可轉(zhuǎn)化為等差、等比數(shù)列的問題可轉(zhuǎn)化
3、為等差、等比數(shù)列的問題例2已知an是公差為3的等差數(shù)列,數(shù)列bn滿足b1=1, b2= ,anbn+1+bn+1=nbn.(1)求an的通項(xiàng)公式;(2)求bn的前n項(xiàng)和.-5-解題心得無論是求數(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.-6-對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練2設(shè)an是公比大于1的等比數(shù)列,Sn為數(shù)列an的前n項(xiàng)和,已知S3=7,且a1+3,3a2,a3+4構(gòu)成等差數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)令bn= ,n=1,2,求數(shù)列bn的前n項(xiàng)和Tn.-7-(2)由(1)得a3n+1=23n,bn
4、=ln 23n=3nln 2.bn+1-bn=3ln 2,數(shù)列bn為等差數(shù)列.-8-求數(shù)列的通項(xiàng)及錯(cuò)位相減求和求數(shù)列的通項(xiàng)及錯(cuò)位相減求和例3已知an為等差數(shù)列,前n項(xiàng)和為Sn(nN*),bn是首項(xiàng)為2的等比數(shù)列,且公比大于0,b2+b3=12,b3=a4-2a1,S11=11b4.(1)求an和bn的通項(xiàng)公式;(2)求數(shù)列a2nbn的前n項(xiàng)和(nN*).解 (1)設(shè)等差數(shù)列an的公差為d,等比數(shù)列bn的公比為q.由已知b2+b3=12,得b1(q+q2)=12,而b1=2,所以q2+q-6=0.又因?yàn)閝0,解得q=2.所以,bn=2n.由b3=a4-2a1,可得3d-a1=8.由S11=11b
5、4,可得a1+5d=16,聯(lián)立,解得a1=1,d=3,由此可得an=3n-2.所以,an的通項(xiàng)公式為an=3n-2,bn的通項(xiàng)公式為bn=2n.-9-(2)設(shè)數(shù)列a2nbn的前n項(xiàng)和為Tn,由a2n=6n-2,有Tn=42+1022+1623+(6n-2)2n,2Tn=422+1023+1624+(6n-8)2n+(6n-2)2n+1,上述兩式相減,得-Tn=42+622+623+62n-(6n-2)2n+1得Tn=(3n-4)2n+2+16.所以,數(shù)列a2nbn的前n項(xiàng)和為(3n-4)2n+2+16.-10-解題心得求數(shù)列通項(xiàng)的基本方法是利用等差、等比數(shù)列通項(xiàng)公式,或通過變形轉(zhuǎn)換成等差、等比
6、數(shù)列求通項(xiàng);如果數(shù)列an與數(shù)列bn分別是等差數(shù)列和等比數(shù)列,那么數(shù)列anbn的前n項(xiàng)和采用錯(cuò)位相減法來求.-11-對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練3(2018浙江,20)已知等比數(shù)列an的公比q1,且a3+a4+a5=28,a4+2是a3,a5的等差中項(xiàng).數(shù)列bn滿足b1=1,數(shù)列(bn+1-bn)an的前n項(xiàng)和為2n2+n.(1)求q的值;(2)求數(shù)列bn的通項(xiàng)公式.解 (1)由a4+2是a3,a5的等差中項(xiàng),得a3+a5=2a4+4,所以a3+a4+a5=3a4+4=28,解得a4=8.-12-13-14-求數(shù)列的通項(xiàng)及裂項(xiàng)求和求數(shù)列的通項(xiàng)及裂項(xiàng)求和例4設(shè)數(shù)列an滿足a1+3a2+(2n-1)an=2n.
7、(1)求an的通項(xiàng)公式;(2)求數(shù)列 的前n項(xiàng)和.-15-解 (1)因?yàn)閍1+3a2+(2n-1)an=2n,故當(dāng)n2時(shí),a1+3a2+(2n-3)an-1=2(n-1).兩式相減得(2n-1)an=2.-16-解題心得對(duì)于已知等式中含有an,Sn的求數(shù)列通項(xiàng)的題目,一般有兩種解題思路,一是消去Sn得到f(an)=0,求出an;二是消去an得到g(Sn)=0,求出Sn,再求an.把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差,求和時(shí)中間的項(xiàng)能夠抵消,從而求得其和.注意抵消后所剩余的項(xiàng)一般前后對(duì)稱.-17-對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練4已知an為公差不為零的等差數(shù)列,其中a1,a2,a5成等比數(shù)列,a3+a4=12.(1)求數(shù)列
8、an的通項(xiàng)公式;-18-19-涉及奇偶數(shù)討論的數(shù)列求和涉及奇偶數(shù)討論的數(shù)列求和例5已知等差數(shù)列an的前n項(xiàng)和為Sn,且a1=2,S5=30.數(shù)列bn的前n項(xiàng)和為Tn,且Tn=2n-1.(1)求數(shù)列an,bn的通項(xiàng)公式;(2)設(shè)cn=(-1)n(anbn+ln Sn),求數(shù)列cn的前n項(xiàng)和.解 (1)S5=5a1+ d=10+10d=30,d=2,an=2n.對(duì)數(shù)列bn:當(dāng)n=1時(shí),b1=T1=21-1=1,當(dāng)n2時(shí),bn=Tn-Tn-1=2n-2n-1=2n-1,當(dāng)n=1時(shí)也滿足上式.bn=2n-1.-20-21-當(dāng)n為偶數(shù)時(shí),Bn=-(ln 1+ln 2)+(ln 2+ln 3)-(ln 3+ln 4)+ln n+ln(n+1)=ln(n+1)-ln 1=ln(n+1);當(dāng)n為奇數(shù)時(shí),Bn=-(ln 1+ln 2)+(ln 2+ln 3)-(ln 3+ln 4)+-ln n+ln(n+1)=-ln(n+1)-ln 1=-ln(n+1).由以上可知,Bn=(-1)nln(n+1).-22-對(duì)點(diǎn)訓(xùn)練對(duì)點(diǎn)訓(xùn)練5已知函數(shù)f(x)=4x,4,f(a1),f(a2),f(an),2n+3(nN*)成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;-23-
高考數(shù)學(xué)二輪復(fù)習(xí) 專題四 數(shù)列 4.2.1 等差、等比數(shù)列與數(shù)列的通項(xiàng)及求和課件 文
