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1、第四節(jié)數(shù)列求和總綱目錄教材研讀1.求數(shù)列的前n項(xiàng)和的方法考點(diǎn)突破2.常見(jiàn)的裂項(xiàng)公式考點(diǎn)二裂項(xiàng)相消法求和考點(diǎn)二裂項(xiàng)相消法求和考點(diǎn)一錯(cuò)位相減法求和考點(diǎn)三分組轉(zhuǎn)化法求和考點(diǎn)三分組轉(zhuǎn)化法求和1.求數(shù)列的前求數(shù)列的前n項(xiàng)和的方法項(xiàng)和的方法(1)公式法公式法(i)等差數(shù)列的前n項(xiàng)和公式Sn=na1+.(ii)等比數(shù)列的前n項(xiàng)和公式當(dāng)q=1時(shí),Sn=na1;1()2nn aa(1)2n nd教材研讀教材研讀當(dāng)q1時(shí),Sn=.(2)分組轉(zhuǎn)化法分組轉(zhuǎn)化法把數(shù)列的每一項(xiàng)轉(zhuǎn)化成幾項(xiàng)之和,使所求和轉(zhuǎn)化為幾個(gè)等差、等比數(shù)列之和,再求解.(3)裂項(xiàng)相消法裂項(xiàng)相消法把數(shù)列的通項(xiàng)拆成兩項(xiàng)之差求和,正負(fù)相消剩下首尾若干項(xiàng).(4

2、)倒序相加法倒序相加法把數(shù)列分別正著寫(xiě)和倒著寫(xiě)再相加,倒序相加法是對(duì)等差數(shù)列求和公式的推導(dǎo)過(guò)程的推廣.1(1)1naqq11naa qq(5)錯(cuò)位相減法錯(cuò)位相減法適用于一個(gè)等差數(shù)列與一個(gè)等比數(shù)列對(duì)應(yīng)項(xiàng)相乘所得的數(shù)列的求和,錯(cuò)位相減法是對(duì)等比數(shù)列求和公式的推導(dǎo)過(guò)程的推廣.(6)并項(xiàng)求和法并項(xiàng)求和法若一個(gè)數(shù)列的前n項(xiàng)和中,可兩兩合并求解,這種方法稱(chēng)為并項(xiàng)求和.形如an=(-1)nf(n)類(lèi)型,可采用兩項(xiàng)合并求解.例如,Sn=1002-992+982-972+22-12=(100+99)+(98+97)+(2+1)=5050.2.常見(jiàn)的裂項(xiàng)公式常見(jiàn)的裂項(xiàng)公式(1)=-;(2)=;(3)=-.1(1)

3、n n1n11n1(21)(21)nn12112121nn11nn1nn1.若數(shù)列an的通項(xiàng)公式為an=2n+2n-1,則它的前n項(xiàng)和Sn=()A.2n+n2-1B.2n+1+n2-1C.2n+1+n2-2D.2n+n2-2答案答案CSn=(21+1)+(22+3)+(23+5)+(2n+2n-1)=(21+22+2n)+1+3+5+(2n-1)=+=2n+1-2+n2.故選C.2(1 2 )1 2n1(21)2nnC2.已知數(shù)列an的前n項(xiàng)和為Sn=1-5+9-13+17-21+(-1)n-1(4n-3),則S15+S22-S31的值是()A.13B.-76C.46D.76答案答案BS15=

4、1-5+9-13+(413-3)-(414-3)+(415-3)=7(-4)+57=29,BS22=1-5+9-13+(421-3)-(422-3)=11(-4)=-44,S31=1-5+9-13+(429-3)-(430-3)+(431-3)=15(-4)+121=61,S15+S22-S31=29-44-61=-76.故選B.3.數(shù)列的前n項(xiàng)之和為,則n=.1(1)n n991003.數(shù)列的前n項(xiàng)之和為,則n=.1(1)n n99100答案答案99解析解析由題意得+=-+-+-+-=1-=,令=,解得n=99.11 212 313 41(1)nn1112121313141n11n11n1n

5、n1nn99100994.已知數(shù)列an的前n項(xiàng)和為Sn,且an=n2n,則Sn=.(n-1)2n+1+2答案答案(n-1)2n+1+2解析解析an=n2n,Sn=121+222+323+n2n.2Sn=122+223+(n-1)2n+n2n+1.-,得-Sn=2+22+23+2n-n2n+1=-n2n+1=2n+1-2-n2n+12(1 2 )1 2n=(1-n)2n+1-2.Sn=(n-1)2n+1+2.典例典例1(2015北京朝陽(yáng)一模)設(shè)數(shù)列an的前n項(xiàng)和為Sn,且a1=4,an+1=Sn,nN*.(1)寫(xiě)出a2,a3,a4的值;(2)求數(shù)列an的通項(xiàng)公式;(3)已知等差數(shù)列bn中,有b2

6、=a2,b3=a3,求數(shù)列anbn的前n項(xiàng)和Tn.考點(diǎn)一錯(cuò)位相減法求和考點(diǎn)一錯(cuò)位相減法求和考點(diǎn)突破考點(diǎn)突破(2)當(dāng)n2時(shí),an=Sn-Sn-1=2n+1-2n=2n.又當(dāng)n=1時(shí),a1=S1=4.所以an=(3)設(shè)等差數(shù)列bn的公差為d,依題意,b2=a2=4,b3=a3=8,則由得b1=0,d=4,則bn=4(n-1).所以anbn=因?yàn)楫?dāng)n=1時(shí),(n-1)2n+2=0,所以anbn=(n-1)2n+2(nN*).4,1,2 ,2.nnn114,28,bdbd20,1,(1)2,2.nnnn解析解析(1)因?yàn)閍1=4,an+1=Sn,所以a2=S1=a1=4,a3=S2=a1+a2=4+4

7、=8,a4=S3=a1+a2+a3=4+4+8=16.所以Tn=a1b1+a2b2+a3b3+a4b4+an-1bn-1+anbn=0+124+225+326+(n-2)2n+1+(n-1)2n+2,2Tn=0+125+226+327+(n-2)2n+2+(n-1)2n+3,-,得-Tn=24+25+26+27+2n+2-(n-1)2n+3=-(n-1)2n+3=-16-(n-2)2n+3.所以Tn=16+(n-2)2n+3.412 (1 2)1 2n方法技巧方法技巧(1)一般地,如果數(shù)列an是等差數(shù)列,bn是等比數(shù)列,求數(shù)列anbn的前n項(xiàng)和時(shí),可采用錯(cuò)位相減法求和,一般是和式兩邊同乘等比數(shù)

8、列bn的公比,然后作差求解;(2)在寫(xiě)出“Sn”與“qSn”的表達(dá)式時(shí)應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對(duì)齊”以便下一步準(zhǔn)確寫(xiě)出“Sn-qSn”的表達(dá)式.1-1已知數(shù)列an是公差大于零的等差數(shù)列,數(shù)列bn為等比數(shù)列,且a1=1,b1=2,b2-a2=1,a3+b3=13.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)設(shè)cn=anbn,求數(shù)列cn的前n項(xiàng)和Tn.解析解析(1)設(shè)數(shù)列an的公差為d(d0),數(shù)列bn的公比為q,由已知得解得或d0,d=2,q=2,an=1+2(n-1)=2n-1,bn=22n-1=2n,即an=2n-1(nN*),bn=2n(nN*).(2)由(1)知cn=anbn=(2n-1)2n

9、,Tn=12+322+523+(2n-1)2n,2Tn=122+323+524+(2n-1)2n+1,22(1)1,12213,qddq10,4dq 2,2.dq-得Tn=-12-222-223-22n+(2n-1)2n+1=-2-23-24-2n+1+(2n-1)2n+1=-2-+(2n-1)2n+1=6+(2n-3)2n+1.312(1 2)1 2n考點(diǎn)二裂項(xiàng)相消法求和考點(diǎn)二裂項(xiàng)相消法求和典例典例2(2016北京東城二模)已知等差數(shù)列an滿(mǎn)足a3=7,a5+a7=26,其前n項(xiàng)和為Sn.(1)求an的通項(xiàng)公式及Sn;(2)令bn=(nN*),求數(shù)列bn的前8項(xiàng)和.1nSn解析解析(1)設(shè)等

10、差數(shù)列an的公差為d,由a5+a7=26,得a6=13,又a6-a3=3d=6,故d=2.所以an=a3+(n-3)d=7+2(n-3)=2n+1.所以Sn=n=n=n2+2n.(2)由bn=,得bn=-.設(shè)bn的前n項(xiàng)和為T(mén)n,則T8=+=1-=.故數(shù)列bn的前8項(xiàng)和為.12naa3212n1nSn21nn1(1)n n1n11n112112311341189198989易錯(cuò)警示易錯(cuò)警示利用裂項(xiàng)相消法求和時(shí),應(yīng)注意抵消后并不一定只剩下第一項(xiàng)和最后一項(xiàng),也有可能前面剩兩項(xiàng),后面也剩兩項(xiàng).有些情況下,裂項(xiàng)時(shí)需要調(diào)整前面的系數(shù),使裂開(kāi)的兩項(xiàng)之差和系數(shù)之積與原通項(xiàng)相等.2-1(2018北京海淀高三期

11、中)已知等比數(shù)列an滿(mǎn)足a1a2a3=8,a5=16.(1)求an的通項(xiàng)公式及前n項(xiàng)和Sn;(2)設(shè)bn=log2an+1,求數(shù)列的前n項(xiàng)和Tn.11nnb b解析解析(1)設(shè)等比數(shù)列an的公比為q.因?yàn)閍1a2a3=8,且a1a3=,所以=8,解得a2=2,又因?yàn)閍5=a2q3=16,所以q3=8,解得q=2,所以a1=1.所以an=2n-1(nN+),所以Sn=2n-1.(2)因?yàn)閍n+1=2n,所以bn=log2an+1=n,所以=-.所以數(shù)列的前n項(xiàng)和Tn=+=1-=.22a32a1(1)1naqq1 21 2n11nnb b1(1)n n1n11n11nnb b1121123111n

12、n11n1nn典例典例3(2017北京西城一模)已知an是等比數(shù)列,a1=3,a4=24.數(shù)列bn滿(mǎn)足b1=1,b4=-8,且an+bn是等差數(shù)列.(1)求數(shù)列an和bn的通項(xiàng)公式;(2)求數(shù)列bn的前n項(xiàng)和.考點(diǎn)三分組轉(zhuǎn)化法求和考點(diǎn)三分組轉(zhuǎn)化法求和解析解析(1)設(shè)等比數(shù)列an的公比為q.由題意得q3=8,解得q=2.所以an=a1qn-1=32n-1.設(shè)等差數(shù)列an+bn的公差為d.由題意得d=4.所以an+bn=(a1+b1)+(n-1)d=4n.從而bn=4n-32n-1(n=1,2,).(2)由(1)知bn=4n-32n-1.設(shè)bn的前n項(xiàng)和為Sn.41aa4411()()4 1aba

13、b1643則Sn=(4+42+4n)-(321-1+322-1+32n-1)=4(1+2+n)-3(20+21+2n-1)=2n(n+1)-3(1 2 )1 2n=2n2+2n+3-32n.所以,數(shù)列bn的前n項(xiàng)和為2n2+2n-32n+3.規(guī)律總結(jié)規(guī)律總結(jié)(1)若an=bncn,且bn,cn為等差或等比數(shù)列,可采用分組轉(zhuǎn)化法求an的前n項(xiàng)和.(2)對(duì)于通項(xiàng)公式為an=的數(shù)列,其中bn,cn是等比數(shù)列或等差數(shù)列,可采用分組轉(zhuǎn)化法求和.(3)采用分組轉(zhuǎn)化法求和是將所求數(shù)列和分解轉(zhuǎn)化為若干個(gè)可求和的新數(shù)列的和或差,從而求得原數(shù)列的和,這就需要通過(guò)對(duì)數(shù)列通項(xiàng)結(jié)構(gòu)特點(diǎn)進(jìn)行分析研究,將數(shù)列的通項(xiàng)合理分解

14、轉(zhuǎn)化.,nnb nc n為奇數(shù)為偶數(shù)3-1(2016北京海淀二模)已知等差數(shù)列an的通項(xiàng)公式為an=4n-2,各項(xiàng)都是正數(shù)的等比數(shù)列bn滿(mǎn)足b1=a1,b2+b3=a3+2.(1)求數(shù)列bn的通項(xiàng)公式;(2)求數(shù)列an+bn的前n項(xiàng)和Sn.解析解析(1)設(shè)數(shù)列bn的公比為q,q0,因?yàn)閎1=a1=2,所以b2+b3=2q+2q2=a3+2=12.解得q=2或q=-3(舍).所以bn=b1qn-1=2n.(2)記an的前n項(xiàng)和為T(mén)n,bn的前n項(xiàng)和為Hn,所以Tn=n=n=2n2,12naa2422nHn=2n+1-2.1(1)1nbqq2(1 2 )1n所以Sn=Tn+Hn=2n2+2n+1-2.