2019年高考數(shù)學(xué)真題分類匯編 6.4 數(shù)列求和、數(shù)列的綜合應(yīng)用 文考點(diǎn)一數(shù)列求和1.(xx課標(biāo),17,12分)已知an是遞增的等差數(shù)列,a2,a4是方程x2-5x+6=0的根.(1)求an的通項(xiàng)公式;(2)求數(shù)列的前n項(xiàng)和.解析(1)方程x2-5x+6=0的兩根為2,3,由題意得a2=2,a4=3.設(shè)數(shù)列an的公差為d,則a4-a2=2d,故d=,從而a1=.所以an的通項(xiàng)公式為an=n+1.(2)設(shè)的前n項(xiàng)和為Sn,由(1)知=,則Sn=+,Sn=+.兩式相減得Sn=+-=+-.所以Sn=2-.2.(xx安徽,18,12分)數(shù)列an滿足a1=1,nan+1=(n+1)an+n(n+1),nN*.(1)證明:數(shù)列是等差數(shù)列;(2)設(shè)bn=3n,求數(shù)列bn的前n項(xiàng)和Sn.解析(1)證明:由已知可得=+1,即-=1.所以是以=1為首項(xiàng),1為公差的等差數(shù)列.(2)由(1)得=1+(n-1)1=n,所以an=n2.從而bn=n3n.Sn=131+232+333+n3n,3Sn=132+233+(n-1)3n+n3n+1.-得-2Sn=31+32+3n-n3n+1=-n3n+1=.所以Sn=.3.(xx山東,19,12分)在等差數(shù)列an中,已知公差d=2,a2是a1與a4的等比中項(xiàng).(1)求數(shù)列an的通項(xiàng)公式;(2)設(shè)bn=,記Tn=-b1+b2-b3+b4-+(-1)nbn,求Tn.解析(1)由題意知(a1+d)2=a1(a1+3d),即(a1+2)2=a1(a1+6),解得a1=2,所以數(shù)列an的通項(xiàng)公式為an=2n.(2)由題意知bn=n(n+1).所以Tn=-12+23-34+(-1)nn(n+1).因?yàn)閎n+1-bn=2(n+1),所以當(dāng)n為偶數(shù)時(shí),Tn=(-b1+b2)+(-b3+b4)+(-bn-1+bn)=4+8+12+2n=,當(dāng)n為奇數(shù)時(shí),Tn=Tn-1+(-bn)=-n(n+1)=-.所以Tn=4.(xx四川,19,12分)設(shè)等差數(shù)列an的公差為d,點(diǎn)(an,bn)在函數(shù)f(x)=2x的圖象上(nN*).(1)證明:數(shù)列bn為等比數(shù)列;(2)若a1=1,函數(shù)f(x)的圖象在點(diǎn)(a2,b2)處的切線在x軸上的截距為2-,求數(shù)列an的前n項(xiàng)和Sn.解析(1)證明:由已知可知,bn=>0,當(dāng)n1時(shí),=2d,所以數(shù)列bn是首項(xiàng)為,公比為2d的等比數(shù)列.(2)函數(shù)f(x)=2x在(a2,b2)處的切線方程為y-=(ln 2)(x-a2),它在x軸上的截距為a2-.由題意知,a2-=2-,解得a2=2.所以d=a2-a1=1,an=n,bn=2n,an=n4n.于是,Sn=14+242+343+(n-1)4n-1+n4n,4Sn=142+243+(n-1)4n+n4n+1,因此Sn-4Sn=4+42+4n-n4n+1=-n4n+1=.所以Sn=.考點(diǎn)二數(shù)列的綜合應(yīng)用5.(xx安徽,12,5分)如圖,在等腰直角三角形ABC中,斜邊BC=2.過點(diǎn)A作BC的垂線,垂足為A1;過點(diǎn)A1作AC的垂線,垂足為A2;過點(diǎn)A2作A1C的垂線,垂足為A3;,依此類推.設(shè)BA=a1,AA1=a2,A1A2=a3,A5A6=a7,則a7=.答案6.(xx廣東,19,14分)設(shè)各項(xiàng)均為正數(shù)的數(shù)列an的前n項(xiàng)和為Sn,且Sn滿足-(n2+n-3)Sn-3(n2+n)=0,nN*.(1)求a1的值;(2)求數(shù)列an的通項(xiàng)公式;(3)證明:對(duì)一切正整數(shù)n,有+<.解析(1)-(n2+n-3)Sn-3(n2+n)=0,令n=1,得+a1-6=0,解得a1=2或a1=-3.又an>0,a1=2.(2)由-(n2+n-3)Sn-3(n2+n)=0,得Sn-(n2+n)(Sn+3)=0,又an>0,所以Sn+30,所以Sn=n2+n,所以當(dāng)n2時(shí),an=Sn-Sn-1=n2+n-(n-1)2+n-1=2n,又由(1)知,a1=2,符合上式,所以an=2n.(3)證明:由(2)知,=,所以+=+<+<+=+<+=.7.(xx湖北,19,12分)已知等差數(shù)列an滿足:a1=2,且a1,a2,a5成等比數(shù)列.(1)求數(shù)列an的通項(xiàng)公式;(2)記Sn為數(shù)列an的前n項(xiàng)和,是否存在正整數(shù)n,使得Sn>60n+800?若存在,求n的最小值;若不存在,說明理由.解析(1)設(shè)數(shù)列an的公差為d,依題意,2,2+d,2+4d成等比數(shù)列,故有(2+d)2=2(2+4d),化簡(jiǎn)得d2-4d=0,解得d=0或d=4.當(dāng)d=0時(shí),an=2;當(dāng)d=4時(shí),an=2+(n-1)4=4n-2,從而得數(shù)列an的通項(xiàng)公式為an=2或an=4n-2.(2)當(dāng)an=2時(shí),Sn=2n.顯然2n<60n+800,此時(shí)不存在正整數(shù)n,使得Sn>60n+800成立.當(dāng)an=4n-2時(shí),Sn=2n2.令2n2>60n+800,即n2-30n-400>0,解得n>40或n<-10(舍去),此時(shí)存在正整數(shù)n,使得Sn>60n+800成立,n的最小值為41.綜上,當(dāng)an=2時(shí),不存在滿足題意的n;當(dāng)an=4n-2時(shí),存在滿足題意的n,其最小值為41.8.(xx湖南,21,13分)已知函數(shù)f(x)=xcos x-sin x+1(x>0).(1)求f(x)的單調(diào)區(qū)間;(2)記xi為f(x)的從小到大的第i(iN*)個(gè)零點(diǎn),證明:對(duì)一切nN*,有+<.解析(1)f (x)=cos x-xsin x-cos x=-xsin x.令f (x)=0,得x=k(kN*).當(dāng)x(2k,(2k+1)(kN)時(shí),sin x>0,此時(shí)f (x)<0;當(dāng)x(2k+1),(2k+2)(kN)時(shí),sin x<0,此時(shí)f (x)>0,故f(x)的單調(diào)遞減區(qū)間為(2k,(2k+1)(kN),單調(diào)遞增區(qū)間為(2k+1),(2k+2)(kN).(2)由(1)知, f(x)在區(qū)間(0,)上單調(diào)遞減,又f=0,故x1=,當(dāng)nN*時(shí),因?yàn)閒(n)f(n+1)=(-1)nn+1(-1)n+1(n+1)n+1<0,且函數(shù)f(x)的圖象是連續(xù)不斷的,所以f(x)在區(qū)間(n,(n+1)內(nèi)至少存在一個(gè)零點(diǎn).又f(x)在區(qū)間(n,(n+1)上是單調(diào)的,故n<xn+1<(n+1).因此當(dāng)n=1時(shí),=<當(dāng)n=2時(shí),+<(4+1)<當(dāng)n3時(shí),+<<<=<<.綜上所述,對(duì)一切nN*,+<.