08數(shù)列求和的方法1.已知數(shù)列5,6,1,-5,該數(shù)列的特點(diǎn)是從第二項(xiàng)起,每一項(xiàng)都等于它的前后兩項(xiàng)之和,則這個(gè)數(shù)列的前16項(xiàng)之和S16等于().A.5B.6C.7D.16解析根據(jù)題意得這個(gè)數(shù)列的前8項(xiàng)分別為5,6,1,-5,-6,-1,5,6,發(fā)現(xiàn)從第7項(xiàng)起,數(shù)字重復(fù)出現(xiàn),所以此數(shù)列為周期數(shù)列,且周期為6,前6項(xiàng)和為5+6+1+(-5)+(-6)+(-1)=0.又因?yàn)?6=26+4,所以這個(gè)數(shù)列的前16項(xiàng)之和S16=20+7=7.故選C.答案C2.已知在等差數(shù)列an中,|a3|=|a9|,公差d<0,Sn是數(shù)列an的前n項(xiàng)和,則().A.S5>S6B.S5<S6C.S6=0D.S5=S6解析d<0,|a3|=|a9|,a3>0,a9<0,且a3+a9=2a6=0,a6=0,a5>0,a7<0,S5=S6.故選D.答案D3.若數(shù)列1n(n+1)的前n項(xiàng)和為99100,則n=.解析由題意得112+123+134+1n(n+1)=1-12+12-13+13-14+1n-1n+1=1-1n+1=nn+1=99100,n=99.答案994.已知等差數(shù)列an的公差d>0,前n項(xiàng)和為Sn,a2,a4是方程x2-10x+21=0的兩個(gè)根.(1)求證:1S2+1S3+1Sn<1.(2)求數(shù)列2-nan的前n項(xiàng)和Tn.解析(1)方程x2-10x+21=0的兩個(gè)根分別為x1=3,x2=7,由題意得a2=3,a4=7,a1+d=3,a1+3d=7,解得a1=1,d=2,an=1+(n-1)2=2n-1,Sn=n1+n(n-1)22=n2.當(dāng)n2時(shí),Sn=n2>n(n-1),1S2+1S3+1Sn<112+123+1n(n-1)=1-12+12-13+1n-1-1n=1-1n<1.(2)2-nan=2n-12n,Tn=121+322+523+2n-12n,12Tn=122+323+2n-32n+2n-12n+1.由-得12Tn=12+2122+123+12n-2n-12n+1=12+21221-12n-11-12-2n-12n+1=32-2n+3212n,Tn=3-2n+32n.能力1會(huì)用分組求和法求和【例1】已知數(shù)列an滿足a1=2,a2+a4=8,且對任意nN*,函數(shù)f(x)=(an-an+1+an+2)x+an+1cos x-an+2sin x滿足f2=0.(1)求數(shù)列an的通項(xiàng)公式;(2)若bn=2an+12an,求數(shù)列bn的前n項(xiàng)和Sn.解析(1)由題設(shè)可得f(x)=an-an+1+an+2-an+1sin x-an+2cos x.對任意nN*,f2=an-an+1+an+2-an+1=0,即an+1-an=an+2-an+1,故數(shù)列an為等差數(shù)列.由a1=2,a2+a4=8,得數(shù)列an的公差d=1,所以an=2+1(n-1)=n+1.(2)因?yàn)閎n=2an+12an=2n+1+12n+1=2n+12n+2,所以Sn=b1+b2+bn=(2+2+2)+2(1+2+n)+12+122+12n=2n+2n(n+1)2+121-12n1-12=n2+3n+1-12n.某些數(shù)列的求和是將數(shù)列轉(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.注意在含有字母的數(shù)列中對字母的討論.(1)若數(shù)列cn的通項(xiàng)公式為cn=anbn,且an,bn為等差數(shù)列或等比數(shù)列,則可以采用分組求和法求數(shù)列cn的前n項(xiàng)和.(2)若數(shù)列cn的通項(xiàng)公式為cn=an,n為奇數(shù),bn,n為偶數(shù),且數(shù)列an,bn是等比數(shù)列或等差數(shù)列,則可以采用分組求和法求數(shù)列cn的前n項(xiàng)和.(3)若數(shù)列的通項(xiàng)式中有(-1)n等特征,根據(jù)正號、負(fù)號分組求和.已知在等比數(shù)列an中,an>0,a1=4,1an-1an+1=2an+2,nN*.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=(-1)n(log2an)2,求數(shù)列bn的前2n項(xiàng)和T2n.解析(1)設(shè)等比數(shù)列an的公比為q,則q>0.由a1=4,1an-1an+1=2an+2,得1a1qn-1-1a1qn=2a1qn+1,解得q=2,所以an=42n-1=2n+1.(2)bn=(-1)n(log2an)2=(-1)n(log22n+1)2=(-1)n(n+1)2.設(shè)cn=n+1,則bn=(-1)ncn2.故T2n=b1+b2+b3+b4+b2n-1+b2n=-c12+c22+(-c32)+c42+(-c2n-12)+c2n2=(-c1+c2)(c1+c2)+(-c3+c4)(c3+c4)+(-c2n-1+c2n)(c2n-1+c2n)=c1+c2+c3+c4+c2n-1+c2n=2n2+(2n+1)2=n(2n+3)=2n2+3n.能力2會(huì)用錯(cuò)位相減法求和【例2】設(shè)數(shù)列an的前n項(xiàng)和為Sn,已知2Sn=3n+3.(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足anbn=log3an,求數(shù)列bn的前n項(xiàng)和Tn.解析(1)因?yàn)?Sn=3n+3,所以當(dāng)n=1時(shí),2a1=3+3,即a1=3;當(dāng)n>1時(shí),2Sn-1=3n-1+3,此時(shí)2an=2Sn-2Sn-1=3n-3n-1=23n-1,即an=3n-1.所以an=3,n=1,3n-1,n>1.(2)因?yàn)閍nbn=log3an,所以當(dāng)n=1時(shí),b1=13;當(dāng)n>1時(shí),bn=31-nlog33n-1=(n-1)31-n.所以T1=b1=13.當(dāng)n>1時(shí),Tn=b1+b2+b3+bn=13+13-1+23-2+(n-1)31-n,所以3Tn=1+130+23-1+(n-1)32-n.兩式相減,得2Tn=23+(30+3-1+3-2+32-n)-(n-1)31-n=23+1-31-n1-3-1-(n-1)31-n=136-2n+123n-1,所以Tn=1312-2n+143n-1.經(jīng)檢驗(yàn),當(dāng)n=1時(shí)也適合上式.故Tn=1312-2n+143n-1.(1)一般地,若數(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,則可以采用錯(cuò)位相減法求和.一般是先將和式兩邊同乘以等比數(shù)列bn的公比,然后作差求解.(2)在寫出“Sn”與“qSn”的表達(dá)式時(shí),應(yīng)特別注意將兩式“錯(cuò)項(xiàng)對齊”,以便下一步準(zhǔn)確寫出“Sn-qSn”的表達(dá)式.(3)在應(yīng)用錯(cuò)位相減法求和時(shí),若等比數(shù)列的公比為參數(shù),應(yīng)分公比等于1和不等于1兩種情況求解.設(shè)數(shù)列an是公差大于0的等差數(shù)列,其前n項(xiàng)和為Sn.已知S3=9,且2a1,a3-1,a4+1構(gòu)成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式; (2)若數(shù)列bn滿足anbn=2n-1(nN*),設(shè)Tn是數(shù)列bn的前n項(xiàng)和,證明:Tn<6.解析(1)設(shè)數(shù)列an的公差為d,則d>0.S3=9,a1+a2+a3=3a2=9,即a2=3.又2a1,a3-1,a4+1成等比數(shù)列,(2+d)2=2(3-d)(4+2d),解得d=2,a1=1,an=1+(n-1)2=2n-1.(2)由anbn=2n-1,得bn=2n-12n-1=(2n-1)12n-1,則Tn=1120+3121+(2n-1)12n-1,12Tn=1121+3122+(2n-3)12n-1+(2n-1)12n,兩式相減得12Tn=1+2121+2122+212n-1-(2n-1)12n=1+1-12n-11-12-2n-12n=3-2n+32n,故Tn=6-2n+32n-1.nN*,Tn=6-2n+32n-1<6.能力3會(huì)用裂項(xiàng)相消法求和【例3】設(shè)數(shù)列an的前n項(xiàng)和為Sn,對任意正整數(shù)n都有6Sn=1-2an.(1)求數(shù)列an的通項(xiàng)公式;(2)若數(shù)列bn滿足bn=log12an,Tn=1b12-1+1b22-1+1bn2-1,求Tn.解析(1)由6Sn=1-2an,得6Sn-1=1-2an-1(n2).兩式相減得6an=2an-1-2an,即an=14an-1(n2).由6S1=6a1=1-2a1,得a1=18.數(shù)列an是等比數(shù)列,公比q=14,an=1814n-1=122n+1.(2)an=122n+1,bn=2n+1,從而1bn2-1=14n(n+1)=141n-1n+1.Tn=141-12+12-13+1n-1n+1=141-1n+1=n4(n+1).(1)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).(2)將通項(xiàng)公式裂項(xiàng)后,有時(shí)候需要調(diào)整前面的系數(shù),使裂開的兩項(xiàng)之差和系數(shù)之積與原通項(xiàng)公式相等.若bn為等比數(shù)列,數(shù)列an滿足:對任意的nN*,有a1b1+a2b2+anbn=(n-1)2n+1+2.已知a1=1,a2=2.(1)求數(shù)列an與bn的通項(xiàng)公式;(2)求數(shù)列1anan+1的前n項(xiàng)和Sn.解析(1)由題意可得a1b1=2,a1b1+a2b2=10,解得b1=2,b2=4,bn=2n.又由題意得,當(dāng)n2時(shí),anbn=(a1b1+a2b2+anbn)-(a1b1+a2b2+an-1bn-1)=n2n,an=n,此式對n=1也成立,數(shù)列an的通項(xiàng)公式為an=n.(2)由(1)可知,1anan+1=1n(n+1)=1n-1n+1,Sn=1-12+12-13+1n-1n+1=1-1n+1=nn+1.能力4會(huì)求等差、等比數(shù)列中關(guān)于絕對值的求和問題【例4】在公差為d的等差數(shù)列an中,已知a1=10,且a1,2a2+2,5a3成等比數(shù)列.(1)求d,an;(2)若d<0,求|a1|+|a2|+|a3|+|an|.解析(1)由題意得a15a3=(2a2+2)2.又由a1=10,an是公差為d的等差數(shù)列,得d2-3d-4=0,解得d=-1或d=4.所以an=-n+11(nN*)或an=4n+6(nN*).(2)設(shè)數(shù)列an的前n項(xiàng)和為Sn.因?yàn)閐<0,所以由(1)得d=-1,an=-n+11,所以當(dāng)n11時(shí),|a1|+|a2|+|a3|+|an|=Sn=-12n2+212n;當(dāng)n12時(shí),|a1|+|a2|+|a3|+|an|=-Sn+2S11=12n2-212n+110.綜上所述,|a1|+|a2|+|a3|+|an|=-12n2+212n,n11,12n2-212n+110,n12.根據(jù)等差數(shù)列的通項(xiàng)公式及d<0,確定an的符號,從而去掉絕對值符號,這需要對n的取值范圍進(jìn)行分類討論.已知數(shù)列an滿足a1=-2,an+1=2an+4.(1)證明:數(shù)列an+4是等比數(shù)列.(2)求數(shù)列|an|的前n項(xiàng)和Sn.解析(1)a1=-2,a1+4=2.an+1=2an+4,an+1+4=2an+8=2(an+4),an+1+4an+4=2,an+4是以2為首項(xiàng),2為公比的等比數(shù)列.(2)由(1)可知an+4=2n,an=2n-4.當(dāng)n=1時(shí),a1=-2<0,S1=|a1|=2;當(dāng)n2時(shí),an0,Sn=-a1+a2+an=2+(22-4)+(2n-4)=2+22+2n-4(n-1)=2(1-2n)1-2-4(n-1)=2n+1-4n+2.又當(dāng)n=1時(shí),也滿足上式,Sn=2n+1-4n+2.一、選擇題1.已知數(shù)列an,bn滿足anbn=1,an=n2+3n+2,則bn的前10項(xiàng)之和為().A.13B.512C.12D.712解析bn=1an=1(n+1)(n+2)=1n+1-1n+2,S10=b1+b2+b3+b10=12-13+13-14+14-15+111-112=12-112=512,故選B.答案B2.若數(shù)列an的通項(xiàng)公式為an=(-1)n-1(4n-3),則它的前100項(xiàng)之和S100等于().A.200B.-200C.400D.-400解析S100=(41-3)-(42-3)+(43-3)-(4100-3)=4(1-2)+(3-4)+(99-100)=4(-50)=-200,故選B.答案B3.已知等差數(shù)列an的公差為d,且an0,d0,則1a1a2+1a2a3+1anan+1可化簡為().A.nda1(a1+nd)B.na1(a1+nd)C.da1(a1+nd)D.n+1a1a1+(n+1)d解析1anan+1=1d1an-1an+1,原式=1d1a1-1a2+1a2-1a3+1an-1an+1=1d1a1-1an+1=na1an+1=na1(a1+nd),故選B.答案B4.根據(jù)科學(xué)測算,運(yùn)載神舟飛船的長征系列火箭,在點(diǎn)火后一分鐘上升的高度為1 km,以后每分鐘上升的高度增加2 km,在達(dá)到離地面240 km高度時(shí)船箭分離,則從點(diǎn)火到船箭分離大概需要的時(shí)間是().A.20分鐘B.16分鐘C.14分鐘D.10分鐘解析設(shè)火箭每分鐘上升的高度組成一個(gè)數(shù)列an,顯然a1=1,而an-an-1=2(n2),所以可得an=1+2(n-1)=2n-1.所以Sn=n(a1+an)2=n2=240,所以從點(diǎn)火到船箭分離大概需要的時(shí)間是16分鐘.故選B.答案B5.數(shù)列112,314,518,7116,(2n-1)+12n,的前n項(xiàng)和Sn的值等于().A.n2+1-12nB.2n2-n+1-12nC.n2+1-12n-1D.n2-n+1-12n解析該數(shù)列的通項(xiàng)公式為an=(2n-1)+12n,則Sn=1+3+5+(2n-1)+12+122+12n=n2+1-12n,故選A.答案A6.設(shè)等差數(shù)列an的前n項(xiàng)和為Sn,Sm-1=13,Sm=0,Sm+1=-15,其中mN*且m2,則數(shù)列1anan+1的前n項(xiàng)和的最大值為().A.24143B.1143C.2413D.613解析由題意可得am=Sm-Sm-1=-13,am+1=Sm+1-Sm=-15,故公差d=am+1-am=-2.由Sm=ma1+m(m-1)d2=0,可得a1-m=-1.又由am=a1+(m-1)d=-13,可得a1-2m=-15.所以a1=13,m=14,an=15-2n,故Tn=1a1a2+1a2a3+1anan+1=1d1a1-1a2+1a2-1a3+1an-1an+1=-12113-113-2n=-126+12(13-2n),可知當(dāng)n=6時(shí),Tn取得最大值613,故選D.答案D7.數(shù)列1,(1+2),(1+2+22),(1+2+22+2n-1),的前n項(xiàng)之和Sn為().A.2n-1B.n2n-nC.2n+1-nD.2n+1-n-2解析an=1+2+22+2n-1=2n-1,Sn=2(2n-1)2-1-n=2n+1-2-n,故選D.答案D8.已知數(shù)列1,1,2,1,2,4,1,2,4,8,1,2,4,8,16,即此數(shù)列第一組是20,接下來第二組的兩項(xiàng)是20,21,再接下來第三組的三項(xiàng)是20,21,22,依此類推.設(shè)Sn是此數(shù)列的前n項(xiàng)和,則S2019=().A.264-58B.264-56C.263-58D.263-56解析該數(shù)列第一組有一項(xiàng),其和為20;第二組有兩項(xiàng),其和為20+21;第n組有n項(xiàng),其和為20+21+2n-1=2n-1.前63組共有63642=2016項(xiàng).S2019=20+(20+21)+(20+21+262)+20+21+22=(21-1)+(22-1)+(263-1)+7 =(21+22+263)-(1+1+1)+7=2(1-263)1-2-63+7=264-58,故選A.答案A9.我國古代數(shù)學(xué)名著九章算術(shù)中,有已知長方形面積求一邊的算法,其方法的前兩步如下:第一步,構(gòu)造數(shù)列1,12,13,14,1n;第二步,將數(shù)列的各項(xiàng)乘以n2,得到一個(gè)新數(shù)列a1,a2,a3,an.則a1a2+a2a3+a3a4+an-1an=().A.n24B.(n+1)24C.n(n-1)4D.n(n+1)4解析由題意知,所得新數(shù)列為1n2,12n2,13n2,1nn2,所以a1a2+a2a3+a3a4+an-1an=n24112+123+134+1(n-1)n=n241-12+12-13+13-14+1n-1-1n=n241-1n=n(n-1)4,故選C.答案C二、填空題10.已知數(shù)列an滿足a1=1,an+1an=2n(nN*),則其前2019項(xiàng)和S2019=.解析依題意,得an+1an=2n,an+1an+2=2n+1,則an+1an+2anan+1=2,即an+2an=2,所以數(shù)列a1,a3,a5,a2k-1,是以a1=1為首項(xiàng),2為公比的等比數(shù)列;數(shù)列a2,a4,a6,a2k,是以a2=2為首項(xiàng),2為公比的等比數(shù)列.則S2019=(a1+a3+a5+a2019)+(a2+a4+a6+a2018)=1-210101-2+2(1-21009)1-2=221010-3=21011-3.答案21011-311.若數(shù)列an是正項(xiàng)數(shù)列,且a1+a2+a3+an=n2+n,則a1+a22+ann=.解析由數(shù)列an是正項(xiàng)數(shù)列,且a1+a2+a3+an=n2+n,可得a1=4,a1+a2+an-1=(n-1)2+(n-1)(n2),由-可得an=2n,即an=4n2,ann=4n(n2).則a1+a22+ann=4(1+2+3+n)=2n2+2n(n2),當(dāng)n=1時(shí),a1=4,上述等式也成立.故a1+a22+ann=2n2+2n.答案2n2+2n12.已知數(shù)列an是等差數(shù)列,Sn為其前n項(xiàng)和,若a1=9,a2Z,SnS4,則nSn的最大值為.解析設(shè)數(shù)列an的公差為d.由題意可得a50,a40,即9+4d0,9+3d0,則-3d-94.又因?yàn)閍2=a1+d為整數(shù),所以dZ,即d=-3,此時(shí)an=9-3(n-1)=12-3n,Sn=-32n2+212n.令f(n)=nSn=-32n3+212n2,nN*,則f(x)=-32x3+212x2,f(x)=-92x2+21x.由f(x)=0,可得x=0或x=143,則f(n)的最大值在f(4),f(5)中取得.又f(4)=72,f(5)=75,所以當(dāng)n=5時(shí),f(n)取得最大值75.答案75三、解答題13.已知數(shù)列an的前n項(xiàng)和為Sn,且Sn=n(n+1)2.(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)Tn為數(shù)列bn的前n項(xiàng)和,其中bn=an+12SnSn+1,求Tn.解析(1)當(dāng)n2時(shí),an=Sn-Sn-1=n(n+1)2-(n-1)n2=n,當(dāng)n=1時(shí),a1=S1=1也符合上式,an=n.(2)(法一)bn=an+12SnSn+1=Sn+1-Sn2SnSn+1=121Sn-1Sn+1,Tn=121S1-1S2+1S2-1S3+1Sn-1Sn+1=121S1-1Sn+1=121-2(n+1)(n+2)=n2+3n2(n+1)(n+2).(法二)bn=n+12n(n+1)2(n+1)(n+2)2=2n(n+1)(n+2)=1n(n+1)-1(n+1)(n+2),Tn=12-16+16-112+1n(n+1)-1(n+1)(n+2)=12-1(n+1)(n+2)=n2+3n2(n+1)(n+2).