高中数学专题复习课件：等差数列 20页

《等差数列》\n一、概念与公式1.定义若数列{an}满足:an+1-an=d(常数),则称{an}为等差数列.2.通项公式3.前n项和公式二、等差数列的性质1.首尾项性质:有穷等差数列中,与首末两项距离相等的两项和相等,即:特别地,若项数为奇数,还等于中间项的两倍,即:a1+an=a2+an-1=a3+an-2=…=2a中.a1+an=a2+an-1=a3+an-2=….an=a1+(n-1)d=am+(n-m)d.Sn=na1+=.n(a1+an)2n(n-1)d2\n特别地,若m+n=2p,则am+an=2ap.2.若p+q=r+s(p、q、r、sN*),则ap+aq=ar+as.3.等差中项如果在两个数a、b中间插入一个数A,使a、A、b成等差差数列,则A叫做a与b的等差中项.4.顺次n项和性质5.已知{an}是公差为d的等差数列a+bA=.2(1)若n为奇数,则Sn=na中且S奇-S偶=a中,=.S奇S偶n+1n-1(2)若n为偶数,则S偶-S奇=.nd2若{an}是公差为d的等差数列,则ak,ak,ak也成等差数列,且公差为n2d.k=2n+13nk=1nk=n+12n\n6.若{an},{bn}均为等差数列,则{man},{mankbn}也为等差数列,其中m,k均为常数.三、判断、证明方法1.定义法;2.通项公式法;3.等差中项法.四、Sn的最值问题二次函数注:三个数成等差数列,可设为a-d,a,a+d(或a,a+d,a+2d)四个数成等差数列,可设为a-3d,a-d,a+d,a+3d.7.若等差数列{an}的前2n-1项和为S2n-1,等差数列{bn}的前2n-1项和为T2n-1,则=.S2n-1T2n-1anbn1.若a1>0,d<0时,满足an≥0,an+1≤0.2.若a1<0,d>0时,满足an≤0,an+1≥0.\n典型例题解:不妨设Q>P,则SQ-SP=aP+1+…+aQ=-.P+QPQaP+1+aQ2则SP+Q==(P+Q)(a1+aP+Q)2(P+Q)(aP+1+aQ)2(P+Q)2PQ=-.1.已知,,成等差数列,求证:,,成等差数列.b1a1c1ca+bbc+aab+c2.等差数列的前n项和为Sn,若SP=,SQ=(PQ),求SP+Q(用P,Q表示).QPPQ3.等差数列的前n项和为Sn,若Sm=Sk(m≠k),求Sm+k.4.等差数列{an}的首项a1>0,前n项和为Sn,若Sm=Sk,m≠k,问n为何值时Sn最大.0n=(m+k为偶数时);或(m+k为奇数时).m+k2m+k+12m+k-12\n5.在等差数列{an}中,已知a1=20,前n项和为Sn,且S10=S15.(1)求前n项和Sn;(2)当n为何值时,Sn有最大值,并求它的最大值.(1)Sn=-(n2-25n);56(2)当且仅当n=12或13时,Sn有最大值,最大值为130.6.已知等差数列{an}的前n项和为Sn,且a2=1,S11=33.(1)求数列{an}的通项公式;(2)设bn=(),且数列{bn}的前n项和为Tn,求证:数列{bn}是等比数列,并求Tn.an12(1)an=n;12(2)Tn=(2+1)(1-2-).n2\n7.已知函数f(t)对任意实数x,y都有:f(x+y)=f(x)+f(y)+3xy(x+y+2)+3,f(1)=1.(1)若t为正整数,试求f(t)的表达式;(2)满足f(t)=t的所有整数t能否构成等差数列?若能构成等差数列,求出此数列;若不能构成等差数列,请说明理由;(3)若t为自然数,且t≥4,f(t)≥mt2+(4m+1)t+3m恒成立,求m的最大值.(1)f(t)=t3+3t2-3(tN*);(3)f(t)≥mt2+(4m+1)t+3mf(t)-t≥m(t2+4t+3)m≤t-1.所求数列为:-3,-1,1或1,-1,-3;(2)f(t)=t3+3t2-3(tZ),f(t)=tt=-3,-1,1,故m的最大值是3.\n8.已知函数f(x)=px2+qx,其中,p>0,p+q>1.对于数列{an},设它的前项和为Sn,且Sn=f(n)(nN*).(1)求数列{an}的通项公式;(2)证明:an+1>an>1;(3)证明:点M1(1,),M2(2,),M3(3,),…,Mn(n,)都在同一直线上.1S12S23S3nSn(1)an=(2n-1)p+q(nN*);(2)an+1-an=2p>0,∴an+1>an>a1=p+q=1;(3)只要证其中任意一点Mr(r,)(r>1,rN*)与点M1(1,)1S1rSr连线的斜率为定值(p)即可.\n1.已知{an}是等差数列.(1)前4项和为21,末4项和为67,且各项和为286.求项数;(2)Sn=20,S2n=38,求S3n;(3)项数为奇数,奇数项和为44,偶数项和为33,求数列的中间项和项数.解:(1)设数列的项数为n,依题意得:∴4(a1+an)=21+67=88.∴a1+an=22.∴由n(a1+an)=2Sn=2286得:(2)∵Sn,S2n-Sn,S3n-S2n成等差数列,∴S3n-S2n+Sn=2(S2n-Sn).a1+a2+a3+a4=21,an-3+an-2+an-1+an=67,且有:Sn=286,a1+an=a2+an-1=a3+an-2=a4+an-3.n=26.故所求数列的项数为26.∴S3n=3(S2n-Sn)=3(38-20)=54.(3)依题意S奇+S偶=Sn,S奇-S偶=a中,Sn=na中.Sn=77,a中=11,Sn=na中.解得:a中=11,n=7.课后练习题\n2.等差数列{an},{bn}中,前n项和分别为Sn,Sn,且=,求.SnSn7n+2n+4a5b5解:∵{an},{bn}是等差数列,∴它们的前n项和是关于n的二次函数,且常数项为0,∴a5=S5-S4=65k,b5=S5-S4=13k.a5b5∴==5.65k13kS9S979+29+4a5b5或======5.a1+a92b1+b92a1+a92b1+b92991365∴可设Sn=kn(7n+2),Sn=kn(n+4),\n3.设{an}是一个公差为d(d0)的等差数列,它的前10项和S10=110,且a1,a2,a4成等比数列.(1)证明:a1=d;(2)求公差d的值和数列{an}的通项公式.(1)证:∵a1,a2,a4成等比数列,∴a22=a1a4.而{an}是等差数列,有a2=a1+d,a4=a1+3d.∴(a1+d)2=a1(a1+3d),整理得d2=a1d.∵d0,∴a1=d.(2)解:∵S10=110,而S10=10a1+45d,∴10a1+45d=110,又由(1)知a1=d,代入上式得:11a1=22.即2a1+9d=22.∴a1=2.∴an=2+(n-1)2=2n.∴d=a1=2.∴公差d的值为2,数列{an}的通项公式为an=2n.\n4.已知数列{an}满足a1=4,an=4-(n≥2),令bn=.(1)求证:数列{bn}是等差数列;(2)求数列{an}的通项公式.an-14an-21(1)证:由已知an+1-2=2-=.4an2(an-2)anan+1-21∴==+.2(an-2)anan-2112∴-=.an+1-21an-2112即bn+1-bn=.12故数列{bn}是等差数列.(2)解:∵{}是等差数列,an-21∴=+(n-1)=.a1-21an-21n212∴数列{an}的通项公式为an=2+.2n∴an=2+.2n\n5.数列{an}的前n项和为Sn=npan(nN*),且a1a2,(1)求常数p的值;(2)证明数列{an}是等差数列.(1)解:当n=1时,a1=pa1,若p=1,则当n=2时有a1+a2=2pa2=2a2.∴a1=a2与a1a2矛盾.∴p1.∴a1=0.∴由a1+a2=2pa2知:(2p-1)a2=a1=0.∵a2a1,∴a20,∴p=.12(2)证:由已知Sn=nan,a1=0.12当n≥2时,an=Sn-Sn-1=nan-(n-1)an-1,1212∴=.an-1ann-1n-2则=,…,=.an-2an-1n-2n-3a2a321∴=n-1.a2an∴an=(n-1)a2.∴an-an-1=a2.故数列{an}是以a1为首项,a2为公差的等差数列.\n6.已知数列{an},anN*,Sn=(an+2)2,(1)求证:{an}是等差数列;(2)若bn=an-30,求数列{bn}的前n项和的最小值.1218(1)证:由an+1=Sn+1-Sn得:8an+1=(an+1+2)2-(an+2)2.(2)解:由已知8a1=8S1=(a1+2)2a1=2,故由(1)知an=4n-2.∴(an+1-2)2-(an+2)2=0.∴(an+1+an)(an+1-an-4)=0.∵anN*,∴an+1+an0.∴an+1-an-4=0即an+1-an=4.∴{an}是等差数列.∴bn=2n-1-30=2n-31.解2n-31<0且2(n+1)-31≥0得:≤n<.231229∵nN*,∴n=15.∴{bn}的前15项为负数,其前15项和T15最小.∵b1=-29,公差d=2,故所求前n项和的最小值为-225.∴T15=15(-29)+1572=-225.\n7.已知等差数列{an}的首项是2,前10项之和是15,记An=a2+a4+a8+…+a2n(nN*),求An及An的最大值.解:设等差数列{an}的公差是d,由已知:a1=2且10a1+45d=15.解得:a1=2d=-.19∴An=a2+a4+a8+…+a2n=na1+d[1+3+7+…+(2n-1)]=na1+d(2+22+23+…+2n-n)=2n-(-n)192-12n2-2=(19n+2-2n+1).19求An的最大值有以下解法:法1:由a1>0,d<0,则有a1>a2>…>ak≥0>ak+1>….由ak=2-(k-1)≥0得k≤19.19由k=2n≤19(nN*)得n≤4.即在数列{a2n}中,a21>a22>a23>a24>0>a25>….∴当n=4时,An的值最大,其最大值为:{An}max=(194+2-24+1)=.19946\n解:求An的最大值有以下解法:法2:若存在nN*使得An≥An+1且An≥An-1,则An的值最大.=(19n+2-2n+1),19∵AnAn≥An+1An≥An-1∴19n+2-2n+1≥19(n+1)+2-2n+219n+2-2n+1≥19(n-1)+2-2n解得:9.5≤2n≤19(nN*)n=4.故取n=4时,An的值最大,其最大值为:{An}max=(194+2-24+1)=.199467.已知等差数列{an}的首项是2,前10项之和是15,记An=a2+a4+a8+…+a2n(nN*),求An及An的最大值.\n8.设{an}为等差数列,Sn为数列{an}的前n项和.已知S7=7,S15=75,Tn为数列{}的前n项和.求Tn.Snn解:设等差数列{an}的公差为d,则Sn=na1+.n(n-1)d2∵S7=7,S15=75,解得:a1=-2,d=1.∴Tn=n2-n.94147a1+21d=7,15a1+105d=75,∴a1+3d=1,a1+7d=5,即∴=a1+(n-1)d=-2+(n-1).Snn1212∵-=,Sn+1n+1Snn1212Snn∴数列{}是等差数列,其首项为-2,公差为.\n9.两个数列{an}和{bn}满足bn=,求证:(1)若{bn}为等差数列,则数列{an}也是等差数列;(2)(1)的逆命题也成立.1+2+…+na1+2a2+…+nan证:(1)由已知得a1+2a2+…+nan=n(n+1)bn.①12∴a1+2a2+…+nan+(n+1)an+1=(n+1)(n+2)bn+1.②12将②式减①式化简得:an+1=(n+2)bn+1-nbn.1212∴an=(n+1)bn-(n-1)bn-1=(n+1)bn-(n-1)(2bn-bn+1).12121212∵{bn}为等差数列,∴bn-1=2bn-bn+1,bn+1-bn为常数.∴an+1-an=(n+2)bn+1-nbn-(n+1)bn+(n-1)(2bn-bn+1)12121212=(bn+1-bn)为常数.32故数列{an}也是等差数列.\n证:(2)(1)的逆命题为:两个数列{an}和{bn}满足:1+2+…+na1+2a2+…+nanbn=,若{an}为等差数列,则数列{bn}也是等差数列.证明如下:∵{an}是等差数列,∴可设an=an+b(a,b为常数).∴nan=an2+bn.∴a1+2a2+…+nan=a(12+22+…+n2)+b(1+2+…+n).1+2+…+na1+2a2+…+nan∵bn==an(n+1)(2n+1)+bn(n+1)n(n+1)12121613=a(2n+1)+b.∴bn+1-bn=a,为常数.23故数列{bn}也是等差数列.\n10.已知数列{an}是等差数列,其前n项和为Sn,a3=7,S4=24.(1)求数列{an}的通项公式;(2)设p,q是正整数,且pq,证明:a1+2d=7且4a1+6d=24.解得:a1=3,d=2.∴an=a1+(n-1)d=3+2(n-1)=2n+1.Sp+q<(S2p+S2q).12故数列{an}的通项公式为an=2n+1.(2)证:由(1)知an=2n+1,∴Sn=n2+2n.(1)解:设等差数列{an}的公差为d,依题意得:∵2Sp+q-(S2p+S2q)=2[(p+q)2+2(p+q)]-(4p2+4p)-(4q2+4q)=-2(p-q)2.又pq,∴2Sp+q-(S2p+S2q)<0.Sp+q<(S2p+S2q).12故