2013summer_hefei_alg_sol

代数不等式综合李潜liqian@常用知识和的重要恒等变换式(1)aA+bB−aB−bA=(a−b)(A−B);(2)nXi=1a i!nXi=1b i!=nXi=1nXj=1a ib j=nXi=1nXj=1a jb i;(3)nXi=1a i!2=nXi=1a2i+2X1 i<j na i a j;(4)X1 i<j n (a i−a j)2=nnXi=1a2i−nXi=1a i!2;(5)X1 i<j n a i a j=nXi=1nXj=ia i a j!=nXj=1jXi=1a i a j!;(6)nXi=1nXj=1a ib j=12nXi=1nXj=1(a i b j+a j b i);(7)a n=a1+n−1Xk=1(a k+1−a k).拉格朗日(Lagrange)恒等式设a1,a2,···,a n和b1,b2,···,b n是给定的实数,则有nX i=1a2i!nXi=1b2i!−nXi=1a ib i!2=X1 i<j n(a i b j−a j b i)2.阿贝尔(Abel)恒等式设m<n,m,n∈N∗,则nX k=m (A k−A k−1)b k=A n b n−A m−1b m+n−1Xk=mA k(b k−b k−1),1⃝式1⃝称为阿贝尔(Abel)和差变换公式.在式1⃝中令A0=0,A k=kXi=1a i(1 k n),得nX k=1a k b k=b nnXk=1a k+n−1Xk=1kXi=1a i!(b k−b k+1).2⃝式2⃝称为阿贝尔(Abel)分部求和公式.均值不等式设a 1,a 2,···,a n 是n 个正实数,记H n =n1a 1+1a 2+···+1a n,G n =n√a 1a 2···a n ,A n =a 1+a 2+···+a nn ,Q n =r a 21+a 22+···+a 2nn.分别称为这n 个数的调和平均,几何平均,算术平均和平方平均.这四个平均值之间有如下关系:H n G n A n Q n ,即调和平均不超过几何平均,几何平均又不超过算术平均,算术平均又不超过平方平均.上述三个不等式当且仅当a 1=a 2=···=a n 时成立.柯西(Cauchy)不等式设a 1,a 2,···,a n 和b 1,b 2,···,b n 是给定的实数,则有(a 1b 1+a 2b 2+···+a n b n )2 (a 21+a 22+···+a 2n )(b 21+b 22+···+b 2n ).式中等号成立的充要条件是数组(a 1,a 2,···,a n )和(b 1,b 2,···,b n )成比例,即存在不全为零的实数λ和µ,使λa i =µb i ,i =1,2,···,n.若约定a i =0时,b i =0,则可说当且仅当b 1a 1=b 2a 2=···=b 2a 2时等号成立.T 2引理对于任何实数a 1,a 2,···,a n 和任何正数b 1,b 2,···,b n ,都有a 21b 1+a 22b 2+···+a 2n b n (a 1+a 2+···+a n )2b 1+b 2+···+b n,当且仅当a 1b 1=a 2b 2=···=a n b n时,等号成立.不等式问题选讲1.对任意x,y,z ∈R ,证明:−32(x 2+y 2+2z 2) 3xy +yz +zx 3+√134(x 2+y 2+2z 2).解答左边不等式等价于3x 2+3y 2+6z 2+6xy +2yz +2zx 0⇔3(x +y )2+2(x +y )z +6z 20⇔3 x +y +13z2+173z 20,当且仅当x +y =z =0时,等号成立.因此左边不等式成立.注意到x ,y 的对称性,设3xy +yz +zx 3x 2+3y 2+my 2+1m z 2+1m z 2+mx 2=12 (3+m )x 2+(3+m )y 2+2mz 2.令3+m =1m ,得m =−3+√132(舍去负值).故3+m =3+√132.因此,3xy +yz +zx 3+√134(x 2+y 2+2z 2).当且仅当x =y ,my =z ,mx =z 即z =3+√132x =3+√132y 时,等号成立.因此右边不等式也成立.综上所述,原不等式成立.2.设x ,y ,z 是大于−1的实数.证明:1+x 21+y +z 2+1+y 21+z +x 2+1+z 21+x +y 22.解答一由已知得1+x 2,1+y 2,1+z 2,1+y +z 2,1+z +x 2,1+x +y 2均大于0.由柯西不等式,得 1+x 21+y +z +1+y 21+z +x +1+z 21+x +y[(1+x 2)(1+y +z 2)+(1+y 2)(1+z +x 2)+(1+z 2)(1+x +y 2)] (1+x 2+1+y 2+1+z 2)2.从而1+x 21+y +z +1+y 21+z +x +1+z 21+x +y (x 2+y 2+z 2+3)2(1+x 2)(1+y +z 2)+(1+y 2)(1+z +x 2)+(1+z 2)(1+x +y 2)=x 4+y 4+z 4+9+2x 2y 2+2y 2z 2+2z 2x 2+6x 2+6y 2+6z 2x 2y 2+y 2z 2+z 2x 2+2(x 2+y 2+z 2)+x 2y +y 2z +z 2x +x +y +z +3=2+x 4+y 4+z 4+3+2x 2+2y 2+2z 2−2(x 2y +y 2z +z 2x )=2+(x 2−y )2+(y 2−z )2+(z 2−x )2+(x −1)2+(y −1)2+(z −1)2x 2y 2+y 2z 2+z 2x 2+2(x 2+y 2+z 2)+x 2y +y 2z +z 2x +x +y +z +32.当且仅当x =y =z =1时,上式等号成立.解答二由已知得1+x 2,1+y 2,1+z 2,1+y +z 2,1+z +x 2,1+x +y 2均大于0,于是1+x 21+y +z 2+1+y 21+z +x 2+1+z 21+x +y 21+x 21+z 2+1+y 22+1+y 21+x 2+1+z 22+1+z 21+y 2+1+x 22=2a 2c +b +2b 2a +c +2c2b +a,其中,a =1+x 22,b =1+y 22,c =1+z 22.由柯西不等式,得a2c +b +b 2a +c +c 2b +a (a +b +c )2a (b +2c )+b (c +2a )+c (a +2b ) 3(ab +bc +ca )3(ab +bc +ca )=1.3.设a,b,c >0.证明:(a −b )2(c +a )(c +b )+(b −c )2(a +b )(a +c )+(c −a )2(b +c )(b +a ) (a −b )2a 2+b 2+c 2.解答一由12(a −2b )2+12(a −2c )2+(b −c )2 0⇒3(a 2+b 2+c 2) 2a 2+2ab +2bc +2ac =2(a +b )(a +c )⇒(a +b )(a +c ) 32(a 2+b 2+c 2).同理,(b +a )(b +c ) 32(a 2+b 2+c 2),(c +a )(c +b ) 32(a 2+b 2+c 2).故(a −b )2(c +a )(c +b )+(b −c )2(a +b )(a +c )+(c −a )2(b +c )(b +a ) 23·(a −b )2+(b −c )2+(c −a )2a +b +c 23·(a −b )2+12(b −c +c −a )2a 2+b 2+c 2=(a −b )2a +b +c .解答二记“X”表示转换对称和.由柯西不等式知X(a −b )2(c +a )(c +b )·X (c +a )(c +b ) (|a −b |+|b −c |+|c −a |)2 (|a −b |+|b −c +c −a |)2=4(a −b )2.而X (c +a )(c +b )=X a 2+3X ab 4X a 2,故X(a −b )2(c +a )(c +b ) 4(a −b )24(a +b +c )=(a −b )2a +b +c .4.已知a ,b ,c 是正实数.求证:(2a +b +c )22a +(b +c )+(2b +c +a )22b +(c +a )+(2c +a +b )22c +(a +b ) 8.解答一对a ,b ,c 乘一个合适的因子可把原问题划归为a +b +c =3(a,b,c >0)的情形,故只须证明(a +3)22a 2+(3−a )2+(b +3)22b 2+(3−b )2+(c +3)22c 2+(3−c )28.令f (x )=(x +3)22x 2+(3−x )2.只须证明f (a )+f (b )+f (c ) 8.注意到f (x )=x 2+6x +93(x 2−2x +3)=13·x 2+6x +9x 2−2x +3=13 1+8x +6x 2−2x +3 =13 1+8x +6(x −1)2+213 1+8x +62 =13(4x +4),则f (a )+f (b )+f (c ) 13(4a +4+4b +4+4c +4)=8.解答二由柯西不等式得[a 2+a 2+(b +c )2](12+12+22) [a +a +2(b +c )]2,即2a2+(b+c)2 23(a+b+c)2,从而,12a2+(b+c)232(a+b+c)2.同理,12b2+(c+a)232(a+b+c)2,12c2+(a+b)232(a+b+c)2.注意到(2a+b+c)2 2a2+(b+c)2+(2b+c+a)22b2+(c+a)2+(2c+a+b)22c2+(a+b)2−8=(2a+b+c)2−1+(a+2b+c)2−1+(a+b+2c)2−1−5=2a2+4ab+4ac2a2+(b+c)2+2b2+4ab+4bc2b2+(c+a)2+2c2+4ac+4bc2c2+(a+b)2−53[(2a 2+4ab+4ac)+(2b2+4ab+4bc)+(2c2+4ac+4bc)]2(a+b+c)2−5=−2[(a−b)2+(b−c)2+(c−a)2](a+b+c)20.故原不等式成立.5.设a,b,c为正实数,求a+3c a+2b+c +4ba+b+2c−8ca+b+3c的最小值.解答一令x=a+2b+c,y=a+b+2c,z=a+b+3c,由此可解得a+3c=2y−x,b=z+x−2y, c=z−y,从而,a+3c a+2b+c +4ba+b+2c−8ca+b+3c=2y−xx+4(z+x−2y)y−8(z−y)z=−17+2yx+4xy+4zy+8yz−17+2√8+2√32=−17+12√2式中等号当且仅当y=√2x,z=2x即b=(1+√2)a,c=(4+3√2)a时成立.因此所求的最小值为−17+12√2.解答二不妨设a+b+c=1,则a+3c a+2b+c +4ba+b+2c−8ca+b+3c=1+2c−b1+b+4b1+c−8c1+2c=−1+2+2c1+b+4b+41+c−41+c+41+2c−4=−5+21+c1+b+41+b1+c−4c(1+c)(1+2c)−5+2√8=41c+3+2c−5+4√2−43+2√2=−17+12√2.当a=3−2√22,b=√2−12,c=√22时,等号成立.因此所求的最小值为−17+12√2.6.设正实数a1,a2,···,a n满足a1+a2+···+a n=1.求证:(a1a2+a2a3+···+a n a1)a1a22+a2+a2a23+a3+···+a na21+a1nn+1.解答一由T2引理及题设得a1 2+a23+···+a n1=a2112+a2223+···+a2nn111223n1.因而只须证明a1a22+a2+a2a23+a3+···+a na21+a1nn+1a1a2+a2a3+···+a na1.由T2引理得a1a22+a2+a2a23+a3+···+a na21+a1=a1a22a1+a1a2+a2a32a2+a2a3+···+a na12a n+a na1a1a2+a2a3+···+a na121+a1a2+a2a3+···+a na1.令t=a1a2+a2a3+···+a na1,则t n.从而只须证t21+tntn+1,而此式等价于t n,证毕.解答二由T2引理得a1a22+a2+a2a23+a3+···+a na21+a1=a1a22a1+a1a2+a2a32a2+a2a3+···+a na12a n+a na1a1a2+a2a3+···+a na121+a1a2+a2a3+···+a na1.只需证nX i=1a i a i+1!nXi=1a ia i+1!2nn+1nXi=1a ia i+1+1!.1⃝由柯西不等式得nX i=1a i a i+1!nXi=1a ia i+1!nXi=1a i!2=1,从而证明式1⃝,只需证n X i=1a ia i+1nn+1nXi=1a ia i+1+1!,而此等价于nX i=1a ia i+1n,这由均值不等式得证.解答三定义a n+1=a1.若nX i=1a i a i+11n,则n X i=1a ia i+1(a i+1+1)nXi=1(a i+1+1)nXi=1a ia i+112!2n,于是,nXi=1a ia i+1(a i+1+1)n2n+1,从而,nXi=1a i a i+1nXi=1a ia i+1(a i+1+1)nn+1.若nX i=1a i a i+1<1n,则nXi=1a i a i+1!nXi=1a ia i+1(a i+1+1)!nXi=1a i(a i+1+1)!nXi=1a i!3=1,又n X i =1a i (a i +1+1)<n +1n ,我们有n X i =1a i a i +1n X i =1a ia i +1(a i +1+1)>n n +1.7.设a ,b ,c ,d 为正实数,满足ab +cd =1,点P i (x i ,y i )(i =1,2,3,4)是以原点为圆心的单位圆周上的四个点.求证:(ay 1+by 2+cy 3+dy 4)2+(ax 4+bx 3+cx 2+dx 1)22a 2+b 2ab +c 2+d 2cd.解答一令u =ay 1+by 2,v =cy 3+dy 4,u 1=ax 4+bx 3,v 1=cx 2+dx 1,则u 2 (ay 1+by 2)2+(ax 1−bx 2)2=a 2+b 2+2ab (y 1y 2−x 1x 2),即x 1x 2−y 1y 2 a 2+b 2−u 22ab.1⃝v 21 (cx 2+dx 1)2+(cy 2−dy 1)2=c 2+d 2+2cd (x 1x 2−y 1y 2),即y 1y 2−x 1x 2 c 2+d 2−v 212cd.2⃝1⃝+2⃝,得0 a 2+b 2−u 2ab +c 2+d 2−v 2cd,即u 2ab +v 21cd a 2+b 2ab +c 2+d 2cd.同理,v 2cd +u 21ab c 2+d 2cd +a 2+b 2ab.由柯西不等式,有(u +v )2+(u 1+v 1)2(ab +cd ) u 2ab +v 2cd +(ab +cd ) u 21ab +v 21cd=u 2ab +v 2cd +u 21ab +v 21cd 2 a 2+b 2ab +c 2+d 2cd.解答二记α=ay 1+by 2+cy 3+dy 4,β=ax 4+bx 3+cx 2+dx 1.由柯西不等式,得[(√ady 1)2+(√bcy 2)2+(√bcy 3)2+(√ady 4)2]·" Éa d 2+ r b c 2+ Éc b 2+r da2# (ay 1+by 2+cy 3+dy 4)2=α2,即α2 (ady 21+bcy 22+bcy 23+ady 24)·a d +bc +c b +b a.同理,β2(adx 24+bcx 23+bcx 22+adx 21)·a d +bc +cb +b a.将以上两式相加,并利用x 2i +y 2i =1(i =1,2,3,4),ab +cd =1得,α2+β2(2ad +2bc ) a d +b c +c b +b a=2(ad +bc ) ab +cd bd +ab +cd ac =2(ad +bc ) 1bd +1ac=2 a 2+b 2ab +c 2+d 2cd.8.设x i 0(i =1,2,···,n ),且n X i =1x 2i +2X1 k<j nÊkjx k x j =1.求n Xi =1x i 的最大值与最小值.解答先求最小值.由n X i =1x i!2=n X i =1x 2i+2X1 k<j nx k x j 1,得n X i =1x i 1,等号成立当且仅当存在i 使得x i =1,x j =0,j =i .因此,n Xi =1x i 的最小值为1.再求最大值.令x k =√k ,有n X k =1ky 2k +2X1 k<j nky k y j =1.1⃝设M =n X k =1x k =n Xk =1√ky k ,令8>>><>>>:y 1+y 2+···+y n =a 1,y 2+···+y n =a 2,······yn=a n ,则1⃝⇔a 21+a 22+···+a 2n =1.令a 2n +1=0,则M =n X k =1√a k −a k +1)=n X k =1√k −n X k =1√k +1=n X k =1√k −n X k =1√k −1a k =n X k =1(√−√k −1)a k .由柯西不等式,得M"n X k =1(√k −√k −1)2#12n Xk =1a 2k!12="n X k =1(√k −√k −1)2#12.等号成立当且仅当a 211=···=a 2k(√k −√k −1)2=···=a 2n (√n −√n −1)2⇔a 21+a 22+···+a 2n 1+(√2−√1)2+···+(√n −√n −1)2=a 2k(√k −√k −1)2⇔a k =√k −√k −1"n X k =1(√k −√k −1)2#12(k =1,2,···,n ).由于a1 a2 ··· a n,从而,y k=a k−a k−1=2√−(√k+1+√k−1)"nXk=1(√−√k−1)2#120,即x k 0.所求最大值为"nXk=1(√k−√k−1)2#12.9.设n 2,x1,x2,···,x n均为实数,且nX i=1x2i+n−1Xi=1x i x i+1=1.对于每一个固定的k(k∈N∗,1 k n),求|x k|的最大值.解答一由已知条件得2nXi=1x2i+2n−1Xi=1x i x i+1=2,即x21+(x1+x2)2+(x2+x3)2+···+(x n−2+x n−1)2+(x n−1+x n)2+x2n=2.1⃝由均值不等式,得rx21+(x1+x2)2+···+(x k−1+x k)2k|x1|+|x1+x2|+···+|x k−1+x k|k|x1−(x1+x2)+···+(−1)k−1(x k−1+x k)|k=|x k| k.即x21+(x2+x3)2+···+(x k−1+x k)2 x2k.2⃝同理有r(x k+x k+1)2+···+(x n−1+x n)2+x2nn−k+1|x k+x k+1|+···+|x n−1+x n|+|x n||x n−(x n−1+x n)+···+(−1)n−k(x k+x k+1)|n−k+1=|x k|n−k+1.即(x k+x k+1)2+···+(x n−1+x n)2+x2nx2kn−k+1.3⃝由2⃝+3⃝及1⃝得,21k+1n−k+1x2k,即|x k |r2k (n +1−k )(k =1,2,···,n ).4⃝由上可知,当且仅当x 1=−(x 1+x 2)=x 2+x 3=···=(−1)k −1(x k −1+x k )及x k +x k −1=−(x k +1+x k +2)=···=(−1)n −k x n 时,式4⃝中的等号成立,即当且仅当x i =x k (−1)i −k i(i =1,2,···,k −1)且x j =x k ·(−1)j −k n +1−jn −k +1(j =k +1,k +2,···,n )时等号成立.综上所述,|x k |max =r2k (n +1−k )n +1.解答二首先,我们注意到n X i =1x 2i +n −1Xi =1x i x i +1=1的左边是二次式,且有平方项,显然x i 与x n +1−i 的地位相同,考虑下式(此式是与题给等式等同的):1=(√a 1x 1+√1−a 2x 2)2+(√a 2x 2+√1−a 3x 3)2+···+(√a k −1x k −1+√1−a k x k )2+(p 1−a n +1−k x k +√a n −k x k +1)2+···+(√1−a 2x n −1+√a 1x n )2+[1−(1−a k )+(1−a n +1−k )]x 2k .考虑到x i x i +1项的系数应为1,所以a 1=1,2√a i √1−a i +1=1,即a 1=1,a i +1=1−14a i(i =1,2,···,n −1).易由数学归纳法得到a i =12·i +1i (i =1,2,···,n ),以及1−a i =12·i −1i(i =1,2,···,n ).由以上各式可得[1−(1−a k )−(1−a n +1−k )]x 2k 1,即得|x k |r2k (n +1−k )n +1,(k =1,2,···,n ).易知,当且仅当√a 1x 1+√1−a 2x 2=√a 2x 2+√1−a 3x 3=···=√a k −1x k −1+√1−a k x k =√1−a n +1−k x k +√a n −k x k +1=···=√1−a 2x n −1+√a 1x n =0时等号成立,这样就可求出相应的x 1,x 2,···,x k −1,x k ,x k +1,···,x n 的值,它们满足题给等式.总之,所求的|x k |的最大值是r n +1(k =1,2,3,···,n −1,n ).10.给定大于3的正数n ,设实数x 1,x 2,···,x n ,x n +1,x n +2满足条件0<x 1<x 2<···<x n <x n +1<x n +2.试求n X i =1x i +1x i! n X j =1x j +2x j +1!n X k =1x k +1x k +2x 2k +1+x k x k +2! nX l =1x 2l +1+x l x l +2x l x l +1!的最小值,并求出使该式达到最小值的所有满足条件的实数组x 1,x 2,···,x n ,x n +1,x n +2.解答(I)记t i =x i +1x i(>1),1 i n +1.题中的式子可写成 n X i =1t i ! n Xi =1t i +1!n X i =1t i t i +1t i+t i +1! n Xi =1(t i +t i +1)!.我们看到,n X i =1t i t i +1i i +1!n X i =1(t i +t i +1)!=n X i =1t i −n Xi =1t 2it i +t i +1! n X i =1(t i +t i +1)!=n X i =1t i ! n X i =1(t i +t i +1)!−n X i =1t 2i t i +t i +1!n X i =1(t i +t i +1)!n Xi =1t i!n Xi =1(t i +t i +1)!−n Xi =1t i√t i +t i +1pt i +t i +1!2= n X i =1t i !2+ n X i =1t i ! n X i =1t i +1!− n X i =1t i!2=n X i =1t i! n X i =1t i +1!.因此,对符合条件的实数组0<x 1<x 2<···<x n <x n +1<x n +2,题中的式子不小于1.(II)上面的推演用到柯西不等式,等号成立的充分必要条件是√t i +t i +1t i √t i +t i +1=d,1 i n,这里,d 为常数.也就是t i +1t i=d −1=c ,1 i n .记t 1=b ,有t j =bc j −1,1 j n +1.相应地有x j +1x j=t j =bc j −1,1 j n +1.记x 1=a >0,有x k =t k −1t k −2···t 1a =ab k −1c(k −1)(k −2)2,2 k n +2.由x 2>x 1,得b =x 21>1.又t j =bc j −1>1,1 j n +1,从而,c >n r 1( j −1r 1,1 j n +1).(III)得到结论:(i)对于符合条件的实数组x 1,x 2,···,x n ,x n +1,x n +2,题中式子的最小值是1.(ii)能使该式达到最小值的符合条件0<x 1<x 2<···<x n <x n +1<x n +2的实数组x 1,x 2,···,x n ,x n +1,x n +2应该是x 1=a ,x k =ab k −1c (k −1)(k −2)2,2 k n +2,其中a >0,b >1,c >n r 1b.11.给定正整数n (n 2),设正整数a i (i =1,2,···,n )满足a 1<a 2<···<a n 以及n X i =11a i 1.求证:对于任意实数x ,有n X 1a 2i +x2!212·1a 1(a 1−1)+x 2.解答根据已知条件nXi=11a i1,根据柯西不等式得n X i=11a2i+x!2nXi=11a i!"nXi=1a i(a2i+x)#nXi=1a i(a2i+x).于是,只需证明n X i=1a i(a2i+x2)212·1a1(a1−1)+x2.为此,可证明更强的不等式n X i=1a i(a2i+x)121a1(a1−1)+x−1a n+1(a n+1−1)+x.其中,a n+1>a n为正整数.令b i=121a1(a1−1)+x−1a i+1(a i+1−1)+x.由a i+1 a i+1,得b i−b i−1=121a1(a1−1)+x2−1a i(a i−1)+x2−121a1(a1−1)+x2−1a i+1(a i+1−1)+x2=121a i(a i−1)+x2−1a i+1(a i+1−1)+x2121a i(a i−1)+x2−1(a i+1)a i+x2=a i(a2i+x2)2−a2ia i(a2i+x2)2(i 2).则b n=b1+nXi=2(b i−b i−1)a1(a21+x2)2+nXi=2a i(a2i+x2)2=nXi=1a i(a2i+x2)2.故n X i=1a i(a2i+x2)2121a1(a1−1)+x2−1a n+1(a n+1−1)+x212·1a1(a1−1)+x2.递推数列及其性质12.已知a1=1,a2=3,a n=4a n−1−a n−2(n 3);b1=1,b2=3,b n=b2n−1+2b n−2,(n 3);c1=1,c n+1=2c n+È3c2n−2.求证:对一切正整数n,有a n=b n=c n.解答易知b3=11.由b n=b2n−1+2b n−2(n 3),得b n b n−2=b2n−1+2,1⃝则b n+1b n−1=b2n+2,2⃝2⃝−1⃝,得b n+1b n−1−b n b n−2=b2n−b2n−1,则b n+1+b n−1b n=b n+b n−2b n−1,于是b n+b n−2b=b n−1+b n−3b=···=b3+b1b=4,即b n =4b n −1−b n −2(n 3),因此,a n =b n .易知c 2=3.由c n +1=2c n +È3c 2n −2得(c n +1−2c n )2=3c 2n −2,即c 2n +1−4c n +1c n +c 2n +2=0,3⃝则c 2n −4c n c n −1+c 2n −1+2=0,4⃝3⃝−4⃝并分解因式,得(c n +1−c n −1)(c n +1+c n −1−4c n )=0.易知数列{c n }单调递增,故c n +1>c n −1,于是由上式可得c n +1−4c n +c n −1=0(n 2),即c n =4c n −1−c n −2,因此,a n =c n .综上,对一切n ∈N ∗,都有a n =b n =c n .13.数列{a n }满足a 0=1,a n +1=7a n +È45a 2n −362,n ∈N .证明:(1)对任意n ∈N ,a n 为正整数;(2)对任意n ∈N ,a n a n +1−1为完全平方数.解答(1)由题设得a 1=5,且{a n }严格单调递增.将条件式变形得2a n +1−7a n =È45a 2n −36.两边平方整理,得a 2n +1−7a n a n +1+a 2n +9=0.1⃝由式1⃝又得a 2n −7a n −1a n +a 2n −1+9=0.2⃝1⃝−2⃝得(a n +1−a n −1)(a n +1+a n −1−7a n )=0.又a n +1>a n −1,有a n +1+a n −1−7a n =0,得a n +1=7a n −a n −1.3⃝由式3⃝及a 0=1,a 1=5可知,对任意n ∈N ,a n 为正整数.(2)将式1⃝配方,得(a n +1+a n )2=9(a n a n −1−1),即a n a n −1−1=an +1+a n −132.4⃝由3⃝得a n +1+a n =9a n −(a n +a n −1),则a n +1+a n ≡−(a n +a n −1)≡···≡(−1)n (a 1+a 0)≡0(mod 3).因此,a n +1+a n3为整数,从而a n a n −1−1是完全平方数.14.设数列{a n }满足:a 1=a 2=1,a n =7a n −1−a n −2,n 3.证明:对每个n ∈N ∗,a n +a n +1+2皆为完全平方数.解答易知数列开初的一些项为:1,1,6,41,281,1926,···.注意到,a 1+a 2+2=22,a 2+a 3+2=32,a 3+a 4+2=72,a 4+a 5+2=182,···.构造数列{x n }:x 1=2,x 2=3,x n =3x n −1−x n −2,n 3.则对每个n ∈N ∗,x n 为正整数.我们来证明:对于每个x ∈N ∗,皆有a n +a n +1+2=x 2n .引理数列{x n }满足,对于每个k ∈N ∗,x k x k +2−x 2k +1=5.引理的证明令f(k)=x k x k+1−x2k+1,则f(k)−f(k−1)=(x k x k+2−x2k+1)−(x k−1x k+1−x2k)=(x k x k+2+x2k )−(x2k+1+x k−1x k+1)=x k(x k+2+x k)−x k+1(x k+1+x k−1)=3x k x k+1−3x k+1x k=0.所以f(k)=f(k−1),于是f(k)=f(k−1)=f(k−2)=···=f(1)=x1x3−x22=5.回到本题,对n归纳,根据数列{a n}定义,a1+a2+2=4=x21,a2+a3+2=9=x22,若结论直至n (n 2)皆已成立,则对于n+1,有a n+1+a n+2+2=(7a n−a n−1)+(7a n+1−a n)+2=7(a n+a n+1+2)−(a n−1+a n+2)−10=7x2n−7x2n−1−10=(3x n)2−x2n−1−2x2n−10=(3x n−x n−1)(3x n+x n−1)−2x2n−10=x n+1(x n+1+2x n−1)−2x2n−10=x2n+1+2(x n−1x n+1−x2n−5)=x2n+1.即在n+1时结论也成立,故本题得证.15.已知数列{c n}满足c0=1,c1=0,c2=2005,c n+2=−3c n−4c n−1+2008(n=1,2,3,···).记a n=5(c n+2−c n)(502−c n−1−c n−2)+4n×2004×501(n=2,3,···).问:对n>2,a n是否均为平方数,说明理由.解答由c n+2−c n=−4(c n+c n−1)+2008知a n=−20(c n+c n−1−502)(502−c n−1−c n−2)+ 4n+1×5012.令d n=c n−251,则d n+2=−3d n−4d n−1,d0=−250,d1=−251,d2=1754.a n= 20(d n+d n−1)(d n−1+d n−2)+4n+1×5012.令w n=d n+d n+1,则w n+2=w n+1−4w n,w1=−501, w2=1503,a n=w n w n−1+4n+1×5012.令w n=501T n,则T n+2=T n+1−4T n,T1=−1,T2=3,定义T0=−1,a n=5012×22(5T n T n−1+4n).设x2−x+4=0的两个根为α,β,则α+β=1,αβ=4,T n−αT n−1=β(T n−1−αT n−1)=···=βn−1(T1−αT0)=−βn−1(α−1),T n−βT n−1=−αn−1(β−1),(T n−αT n−1)(T n−βT n−1)= (αβ)n−1(α+1)(β+1)=4n.即T2n−T n T n−1+4T2n−1=4n.从而,a n=5012×22(5T n T n−1+T2n−T n T n−1+4T2n−1)=10022(T n+2T n−1)2.16.已知数列{x n}满足x1=12,x n=Èx n−1+4x n−1+x n−12(n=2,3,···).若y n=nXi=11x2i(n=1,2,···),证明:数列{y n}有极限,并求出该极限.解答因为x n=Èx n−1+4x n−1+x n−12>x n−1,所以,x n>x1=12.故x2n−1+4x n−1>(x n−1+1)2⇒x n>x n−1+12⇒limn→+∞x n=+∞.又x n=Èx n−1+4x n−1+x n−12⇒2x n−x n−1=Èx2n−1+4x n−1⇒4x2n+x2n−1−4x n x n−1=x2n−1+4x n−1⇒1x2n=1x n−1−1x n.故y n=nXi=11x2i=1x1−1x n+4=6−1x n.从而,lim n→+∞y n=limn→+∞6−1x n=6.17.设正数列x1,x2,···,x n,···满足(8x2−7x1)x71=8及x k+1x k−1−x2k=x8k−1−x8k(x x),k 2.求正实数a,使得当x1>a时,有单调性x1>x2>···>x n>···;当0<x1<a时,不具有单调性.解答由x k+1x k−1−x2k=x8k−1−x8kk k−1,有x k+1x k−x kx k−1=1x8k−1x8k−1,即x k+1 x k −1x8k=x kx k−1−1x8k−1=···=x2x1−1x81=78.于是,x k+1=78x k+x−7k,则当x1>0时,x k>0,k 2.由x k+1−x k=x kx−8k−18,则当x−8k−18<0,即x k>818,有x k+1−x k<0,即x k+1<x k,k 1.而x k+1=7x k+x−7k88r1=818,且当x k=818时,等号成立.于是,取a=818,则当x1>818时,x1>x2>···>x n>···.当x1<818时,x2>x1且x2>x3>···>x n.故所求的常数a=818.18.设{a n}为一整数数列,其中,a2为奇数.对任意的正整数n,均有n(a n+1−a n+3)=a n+1+a n+3.已知a2009可被2010整除.求最小的整数n(n 2),使a n可被2010整除.解答将题设等式整理,得(n−1)a n+1=(n+1)a n−3(n−1).当n=1时,a n+1=n+1n−1a n−3⇒a n+1=n(n+1)n(n−1)a n−3.上式两边同时除以n(n+1),得a nn(n+1)=a nn(n−1)−3n(n+1).设b n=a nn(n−1)(n 2).易得b n+1=b n−3n(n+1).故b n=b n−1−3n(n−1),b n−1=b n−2−3(n−1)(n−2),······,b3=b2−12.以上等式求和,得b n=b2−3nXk=31k(k−1)=b2−312−1n.注意到a n=n(n−1)b n,a2=2b2.则a n=n(n−1)a22−312−1n=(n−1)h(a2−3)n2+3i.又a2为奇数,不妨设a2=2p+1(p∈N),则a n=(n−1)[(p−1)n+3].从而,a2009=2008(2009p−2006).又2010|a2009,则2008(2009p−2006)≡0(mod2010)⇒2009p−2006≡0(mod1005)⇒p≡4 (mod1005).从而,存在整数q使得a n=(n−1)[(4+1005q−1)n+3]=(n−1)[(1005q+3)n+3].由2010|a n,得2010|(n−1)[(1005q+3)n+3],即2010|(n−1)(3n+3)⇒670|(n−1)(n+1).故n为奇数,不妨设n=2r+1(r∈N).因此,335|r(r+1).所以,r min=134,n=269.19.数列{a n }定义如下:a 1=1,a n +1=d (a n )+c,n =1,2,···.其中c 是一个确定的正整数,d (m )表示m 的正约数的个数.证明:存在正整数k 使得数列a k ,a k +1,···是周期数列.解答先证明如下引理.引理对任意正整数m ,均有d (m ) m2+1.引理的证明显然,在h m 2i+1到m −1之间不存在m 的约数,则m −d (m ) (m −1)− h m i +1 +1 m−1.从而,d (m ) m2+1.引理得证.下面证明,对任意正整数n (n 2),均有a n 2c +1.用反证法.假设t (t 2)是使得a n 2c +2成立的最小正整数,则d (a t −1)+c 2c +2,即d (a t −1)c +2.又由d (a t −1) a t −12+1,得a t −12+1 c +2,即a t −1 2c +2,这与t 的最小性矛盾.从而,对任何正整数i ,均有a i ∈{1,2,···,2c +1},于是,必存在i ,j (i =j ),使得a i =a j .根据递推数列的性质,知数列{a n }必从某一项起是周期数列.20.设c 是一个正整数.数列a 1,a 2,···,a n ,···,按如下方式定义:a 1=c,a n +1=a 2n +a n +c 3(n ∈N ∗).求c 的一切可能值,使得存在正整数k 1和m 2,满足a 2k +c 3是一个正整数的m 次幂.解答注意到a 2n +1+c 3=(a 2n +a n +c 3)2+c 3=(a 2n +c 3)(a 2n +2a n +1+c 3).先证明:a 2n +c 3与a 2n +2a n +1+c 3互质.注意到(a 2n +3,a 2n +2a n +1+c 3)=(a 2n +c 3,2a n +1).假设存在质数p 3,使得p |(a 2n +c 3),p |(2a n +1).则p |(4a 2n +4c 3)⇒p |[(2a n −1)(2a n +1)+4c 3+1]⇒p |(4c 3+1),于是p |(4c 3+1),p |(2a n +1).而4a n +2=(2a n −1+1)2+4c 3+1,则p |(2a n −1+1)2⇒p |(2a n −1+1).不断递推下去得到p |(2a 1+1).而a 1=c ,从而p |(2c +1),p |(4c 3+1).注意到2(4c 3+1)=(2c +1)(4c 2−2c +1)+1.于是p |1,矛盾.故a 2n +c 3与a 2n +2a n +1+c 3互质.若S m =a 2k +c 3=(a 2k −1+c 3)(a 2k −1+2a k −1+1+c 3),由于a 2k −1+c 3与a 2k −1+2a k −1+1+c 3互质,从而a 2k −1+c 3也是一个数的m 次幂.不断递推下去知,a 21+c 3也是一个整数的m 次幂,即c 2(c +1)=t m .由于(c,c +1)=1,若m 是奇数,则c =t m 1,c +1=t m2,这不可能.若m 为偶数,则c 2(c +1)=t 2m 0⇒c =t m 01,c +1=t 2m 02,于是t m 01+1=t 2m 02⇒(t 22)m 0−t m 01=1,从而必有m 0=1,t 1=t 22−1,则c =r 2−1(r 2,r ∈N +).当c =r 2−1时,a 21+c 3=r 2(r 2−1)2=[r (r 2−1)]2.因此,存在k ,m ,满足a 2k +c 3是一个正整数的m 次幂.21.设两正数列{a n },{b n }满足(1)a 0=1 a 1,a n (b n −1+b n +1)=a n −1b n −1+a n +1b n +1(n 1);(2)n X i =0b i n 32(n 1).求{a n }的通项.解答由条件(1)有a n −a n +1=b n −1b n +1(a n −1−a n ).于是,a n −a n +1=b 0b 1b b (a 0−a 1).1⃝若a1=a0=1,则a n=1.下设a1<a0=1.由式1⃝有a0>a0−a n=b0b1(a0−a1)n−1Xk=01b k b k+1.所以,n X k=01b k b k+1<a0b0b1(a0−a1).2⃝另一方面,令x n=pb n b n+1,则x nb n+b n+12,故nXi=1x inXi=1b i+b i+12<n+1Xi=1b i (n+1)32 4n32.对任意正整数,有k1 x2k+1+1x2k+2+···+1x22kkÈx2k+1 (x2)2kxk+1+ (x2)k2<[4(2k)32]2k2.从而,1 x2k+1+1x2k+2+···+1x22k127.故2n X k=01b k b k+1=2nXk=01x2k=n−1Xk=02kXi=11x22k+1n−1Xk=012=n2,这与式2⃝矛盾.综上所述,a n=1(n 0).。