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7.11数列讲义

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7.11数列讲义 êÆc�šŽùŒ ê�nܯK o d 2010.07 1. ê�{an}½ÂXe: a1 = 1, an+1 = d(an) + c, n = 1, 2, · · · . Ù¥c´˜‡(½���ê, d(m)L«m���ê�‡ê. y²: 3��êk¦�ê�ak, ak+1, · · ·´±Ïê�. )‰ ky²XeÚn. Ún é?¿��êm, þk d(m) ≤ m 2 + 1. Ún�y² w,, 3 [m 2 ] + 1�m− 1ƒmØ3m��ê, K m− d(m) ≥ (m...

7.11数列讲义
êÆc�šŽùŒ ê�nܯK o d 2010.07 1. ê�{an}½ÂXe: a1 = 1, an+1 = d(an) + c, n = 1, 2, · · · . Ù¥c´˜‡(½���ê, d(m)L«m���ê�‡ê. y²: 3��êk¦�ê�ak, ak+1, · · ·´±Ïê�. )‰ ky²XeÚn. Ún é?¿��êm, þk d(m) ≤ m 2 + 1. Ún�y² w,, 3 [m 2 ] + 1�m− 1ƒmØ3m��ê, K m− d(m) ≥ (m− 1)− ([m 2 ] + 1 ) + 1 ≥ m 2 − 1. l , d(m) ≤ m 2 + 1. Ún�y. e¡y², é?¿��ên (n ≥ 2) , þkan ≤ 2c+ 1. ^‡y{. b�t (t ≥ 2)´¦�an ≥ 2c+2¤á����ê,Kd(at−1)+ c ≥ 2c+2,=d(at−1) ≥ c+ 2. qdd(at−1) ≤ at−1 2 + 1, � at−1 2 + 1 ≥ c+ 2, =at−1 ≥ 2c+2, ù†t��5gñ. l , é?Û��êi, þkai ∈ {1, 2, · · · , 2c+1}, u´, 73i, j (i 6= j) , ¦�ai = aj . Šâ4íê��5Ÿ, ê�{an}7l,˜‘å´±Ïê�. 2. �ê�{an}˜‡�ê�(=ˆ‘þ�ê�ê�) , ÷vé?¿��ên, þk (n− 1)an+1 = (n+ 1)an − 2(n− 1). e2000 | a1999, ¦Ñ����ên (n > 1) , ¦�2000 | an. )‰ w,, a1 = 0, an+1 = n+ 1 n− 1an − 2, n ≥ 2. -bn = an n− 1 , Knbn+1 = (n + 1)bn − 2, ½ =n(bn+1 − 2) = (n+ 1)(bn − 2). u´, n ≥ 2ž, þkbn+1 − 2 = n+ 1 n (bn − 2), K bn − 2 = n n− 1 · n− 1 n− 2 · · · · · 3 2 (b2 − 2) = ( b2 2 − 1 ) n, l an = (n − 1) ((a2 2 − 1 ) n+ 2 ) , n ≥ 2. d^‡an ∈ Z, Œa2 2 ∈ Z. �a2 = 2k, Òkan = (n − 1)[(k − 1)n + 2], |^2000 | a1999Œ, 1998(1999k − 1997) ≡ 0 (mod 2000). K−2(−k + 3) ≡ 0 (mod 2000), l k ≡ 3 (mod 1000). ù�, Œ�an = (n− 1)[(1000m+ 2)n+ 2], ùpm ∈ Z, m~ê, n ≥ 2. 1 ddŒ, 2000 | an�¿‡^‡´1000 | (n − 1)(n + 1). l ‡¦nÛê. �n = 2l − 1, K250 | l(l + 1). u(l, l + 1) = 1, 250 = 53 × 2, ¤±k + 1 ≥ 125, n ≥ 2× 124 + 1 = 249. nþŒ, ¤¦�÷v^‡����ên = 249. 3. 鉽���êa, b, b > a > 1, aØU�Øb9‰½���êê�{bn}∞n=1, ÷vé¤k�� ênkbn+1 ≥ 2bn. ´Äo3��ê�ê�{an}∞n=1¦�é¤k��ên, kan+1 − an ∈ {a, b}, … é¤k��êm, l (Œ±ƒÓ) , kam + al 6∈ {bn}∞n=1? )‰ ‰Y´’½�, ·‚^8B{y²{an}�3. Äk�a1��ê, ¦�2a1 6∈ {bn}∞n=1…a1 > b − a (~Xdbn → +∞3n0 ∈ N+, ¦bn0 > b− a+ 1, �a1 = bn0 − 1, Ka1 > b− a…2a1 = 2bn0 − 2 < bn0+1, �2a1 6∈ {bn}∞n=1) . Ùg, b�a1, a2, · · · , ak®�½, ÷vai+1 − ai ∈ {a, b} (i = 1, 2, · · · , k − 1) …al + am 6∈ {bn}∞n=1 (1 ≤ l ≤ k, 1 ≤ m ≤ k) . Ϗak+1 − ak ∈ {a, b}, �ak+1��Škak + a†ak + bü«ŒU, l a1 + ak+1, a2 + ak+1, · · · , ak + ak+1, ak+1 + ak+1��Ške�ü«ŒU: (I) a1 + (ak + a), a2 + (ak + a), · · · , ak + (ak + a), 2(ak + a); (II) a1 + (ak + b), a2 + (ak + b), · · · , ak + (ak + b), 2(ak + b). e¡·‚y²(I) (II) ¥–�k˜‡Ø¹{bn}∞n=1¥�‘. e(I) ¥k{bn}∞n=1¥�˜‘bu, ¿…(II) ¥k{bn}∞n=1¥�˜‘bv, Kdb < a1 + a, �2(ak + b) < 2(a1 + ak + a) ≤ 2bu, l bv ≤ 2(ak + b) < 2bu ≤ bu+1. qbv+1 ≥ 2bv ≥ 2(a1 + ak + b) > 2(a1 + ak + a) > 2(ak + a) ≥ bu, �bu = bv. qbv = bu < 2(ak + a) < 2(ak + b), l 31 ≤ j ≤ k, ¦�bv = aj + ak + b. œ/1: bu = ai + ak + a (1 ≤ i ≤ k) , dž, 0 = bu − bv = ai − aj + a− b, l ai − aj = b− a > 0, d8Bb�, kai − aj = ca+ db (c, dšK�ê) , u´ca+ db = b− a, (1− d)b = (1 + c)a > 0, ¤ ±d = 0, b = (1 + c)a, ù†a - bgñ. œ/2: bu = 2(ak + a), dž0 = bu − bv = ak − aj + 2a − b, l ak − aj = b − 2a, d1 ≤ j ≤ k, ak − aj ≥ 0. 2d8Bb�ak − aj = c′a + d′b (c′, d′šK�ê) , u´c′a + d′b = b − 2a, (1− d′)b = (2 + c′)a > 0, ¤±d′ = 0, b = (2 + c′)a, gñ. �(I) (II)¥–�k˜‡Ø¹{bn}∞n=1¥�‘.ÏdŒ�ak+1 = ak+a½ak+1 = ak+b,¦�ai+ak+1 6∈ {bn}∞n=1 (1 ≤ i ≤ k + 1) , u´·K�y. 4. �p0´‰½�۟ê, ê�a0, a1, a2, · · ·÷va0 = 0, a1 = 1, …an+1 = 2an + (a − 1)an−1, n = 1, 2, · · · . Ù¥a´��ê. ¦÷v±eü‡^‡�a��Š: (i) XJp´Ÿê, …p ≤ p0, Kp | ap; (ii) XJp´Ÿê, …p > p0, Kp - ap. )‰ k4í'X9a0 = 0, a1 = 1Œ� an = 1 2 √ a [(1 + √ a)n − (1−√a)n]. l é?Û۟êp, ap = 1 2 √ a [ p∑ k=0 Cpka k 2 − p∑ k=0 (−1)kCpka a 2 ] = 1 2 √ a ( 2p √ a+ 2C3pa 3 2 + · · ·+ 2Cp−2p a p−2 2 + 2Cppa p 2 ) = p+C3pa+ · · ·+Cp−2p a p−3 2 + a p−1 2 . 2 Ckp = p! k!(p− k)! , u´éu¤k1 ≤ k ≤ p − 1, kp | C p k. ddŒ�, p | ap ⇔ p | a p−1 2 ⇔ p | a. 2 da2 = 2, Œ¤¦a��Š´[3, p0]¥¤kŸê�¦È. 5. Á�ä´Ä3áõ‡4ABC, ÷vAB, BC, CA�´¤��ê��pŸ���ê(AB < BC < CA) , …BC>þ�p94ABC�¡Èþ��ê. ¿‰Ñy². )‰ ·‚òy², 3áõ‡÷v^‡�4ABC. �AB = a− d, BC = a, CA = a+ d, Ù¥a, d ∈ N+, a > d, 4ABC�¡ÈS, BC>þ�pha, Šâ°Ô-‹Ê�úª, � S = √ 3a 2 · (a 2 + d ) · a 2 · (a 2 − d ) = 1 2 √ 3 [(a 2 )2 − d2 ] . eS��ê, Ka7óê, -a = 2x(x ∈ N+), KS = x√3(x2 − d2), u´ ha = 2S a = √ 3(x2 − d2), =h2a = 3(x 2 − d2), l , ha73��ê. �ha = 3y (y ∈ N+) , Kx2 − 3y2 = d2, ؔ�d = 1, K x2 − 3y2 = 1, (1) u´(x, y) = (2, 1)´§(1) �˜‡). -(2 + √ 3)n = xn + yn √ 3 (ùp, xnÚynþ��ê, n ∈ N+) , K(2 − √3)n = xn − yn √ 3, u´x2n − 3y2n = 1, =(x, y) = (xn, yn) (n ∈ N+) Ñ´(1) ���ê). dxn+1 + yn+1 √ 3 = (2 + √ 3)n+1 = (2 + √ 3)(xn + yn √ 3) = (2xn + 3yn) + (xn + 2yn) √ 3, �{ xn+1 = 2xn + 3yn, yn+1 = xn + 2yn, …x1 = 2, y1 = 1. ž�yn, �xn+2 = 4xn+1 − xn, x1 = 2, x2 = 7. u´Œ�AB = 2xn − 1, BC = 2xn, CA = 2xn + 1, n = 1, 2, · · · , KAB, BC, CA´¤��ê��pŸ���ê(AB < BC < CA) , …4ABC�¡ÈS = xn √ 3(x2n − 1) = 3xnyn, BC>þ�pha = 3ynþ��ê. Ïd, 3áõ‡ ÷v^‡�4ABC. 6. ®ê�a1 = 20, a2 = 30, an+2 = 3an+1 − an (n ≥ 1) . ¦¤k���ên, ¦�1 + 5anan+1´� �²ê. )‰ �bn = an + an+1, cn = 1 + 5anan+1, K5an+1 = bn+1 + bn, an+2 − an = bn+1 − bn, u´, cn+1− cn = 5an+1(an+2− an) = b2n+1− b2n. l , cn+1− b2n+1 = cn− b2n = · · · = c1− b21 = 501 = 3× 167. �3��ên, m, ¦�cn = m2, Kk(m+ bn)(m− bn) = 3× 167. em+ bn = 167, m− bn = 3, Km = 85, cn = 852. em+ bn = 501, m− bn = 1, Km = 251, cn = 2512. ê�{an}î‚4O,Kê�{cn}î‚4O,qc1 = 1+ 5× 20× 30 < 852 < c2 = 1+ 5× 30× 70 < c3 = 1 + 5× 70× 180 = 2512. Ïd, ÷v^‡�nk˜‡, =n = 3. 7. ®ê�{cn}÷vc0 = 1, c1 = 0, c2 = 2005, cn+2 = −3cn − 4cn−1 + 2008 (n = 1, 2, 3, · · · ) . Pan = 5(cn+2 − cn)(502− cn−1 − cn−2) + 4n × 2004× 501 (n = 2, 3, · · · ) . ¯: én > 2, an´Äþ² ê, `²nd. )‰ dcn+2 − cn = −4(cn + cn−1) + 2008an = −20(cn + cn−1 − 502)(502 − cn−1 − cn−2) + 4n+1 × 5012. -dn = cn − 251, Kdn+2 = −3dn − 4dn−1, d0 = −250, d1 = −251, d2 = 1754. an = 3 20(dn + dn−1)(dn−1 + dn−2) + 4n+1 × 5012. -wn = dn + dn+1, Kwn+2 = wn+1 − 4wn, w1 = −501, w2 = 1503, an = wnwn−1 + 4n+1 × 5012. -wn = 501Tn, KTn+2 = Tn+1 − 4Tn, T1 = −1, T2 = 3, ½ ÂT0 = −1, an = 5012 × 22(5TnTn−1 + 4n). �x2 − x + 4 = 0�ü‡Šα, β, Kα + β = 1, αβ = 4, Tn − αTn−1 = β(Tn−1 − αTn−1) = · · · = βn−1(T1 − αT0) = −βn−1(α − 1), Tn − βTn−1 = −αn−1(β − 1), (Tn − αTn−1)(Tn − βTn−1) = (αβ)n−1(α+ 1)(β + 1) = 4n. =T 2n − TnTn−1 + 4T 2n−1 = 4n. l , an = 5012 × 22(5TnTn−1 + T 2n − TnTn−1 + 4T 2n−1) = 10022(Tn + 2Tn−1)2. 8. ê�{an}Ú{bn}÷va0 = b0 = 1, an+1 = αan + βbn, bn+1 = βan + γbn, Ù¥α, β, γ��ê, α < γ, …αγ = β2 + 1. y²: é?¿šK�ên, an + bn7ŒLü‡��ê�²Ú. )‰ ky²ü‡Ún. Ún1 e��êm, n, k÷vmn = k2 + 1, K3x1, x2, y1, y2 ∈ N, ¦±enªÓž¤á: m = x21 + y 2 1 , n = x 2 2 + y 2 2 , k = x1x2 + y1y2. Ún1�y² ؔ�m ≤ n, ék^êÆ8B{. �k = 1ž, dmn = 2, km = 1, n = 2, Km = 02 +12, n = 12 +12, k = 0× 1+ 1× 1, =k < 2ž( ؤá. ��k < r (r ≥ 2) ž(ؤá. �k = rž, d mn = r2 + 1, (1) Kn ≥ r + 1, m = r 2 + 1 n ≤ r 2 + 1 r + 1 < r2 + r r + 1 = r. -n = r + s, m = r − t (s, t ∈ N+) , ª(1) ¤(r − t)(r + s) = r2 + 1, =rs− ts− tr = 1. ü>Ó \t2, �(r − t)(s− t) = t2 + 1. r − t = m > 0, Ks− t > 0, t < r. d8Bb�, 3m1,m2, n1, n2 ∈ N, ¦r − t = m21 + n21, s − t = m22 + n22, t = m1m2 + n1n2. =m = r − t = m21 + n21, n = r + s = (r − t) + (s− t) + 2t = (m1 +m2)2 + (n1 + n2)2, r = (r − t)− t = m1(m1 +m2) + n1(n1 + n2). ePx1 = m1, y1 = n1, x2 = m1 +m2, y2 = n1 + n2, K3ª(1) ¥km = x21 + y 2 1 , n = x 2 2 + y 2 2 , k = x1x2 + y1y2 (x1, x2, y1, y2 ∈ N) . Ïd, �k = rž(ؤá. dêÆ8B{, y�Ún1¤á. Ún2 é?¿�m,n ∈ N, k am+n + bm+n = aman + bmbn. (2) Ún2�y² �½Cþm+ n, Iyéu÷v0 ≤ k ≤ m+ n�z‡�êk, k am+n + bm+n = akam+n−k + bkbm+n−k. (3) ék^êÆ8B{. �k = 0ž, (Øw,¤á. �k = 1ž, duam+n = αam+n−1 + βbm+n−1, bm+n = βam+n−1 + γbm+n−1, Kam+n + bm+n = (α+ β)am+n−1 + (β + γ)bm+n−1 = a1am+n−1 + b1bm+n−1. ��k = pž, am+n+bm+n = apam+n−p+bpbm+n−p. �k = p+1ž,kam+n+bm+n = apam+n−p+ bpbm+n−p = ap(αam+n−p−1 + βbm+n−p−1) + bp(βam+n−p−1 + γbm+n−p−1) = (αap + βbp)am+n−p−1 + (βap + γbp)bm+n−p−1 = ap+1am+n−p−1 + bp+1bm+n−p−1. l ª(3) ¤á. 3ª(3) ¥-k = m, Kam+n + bm+n = aman + bmbn. 4 £��K, 3ª(2) ¥, -m = n, Ka2n + b2n = a2n + b 2 n. �m = n + 1, dÚn1ka2n+1 + b2n+1 = an+1an + bn+1bn = (αan + βbn)an + (βan + γbn)bn = αa2n+γb 2 n+2βanbn = (x 2 1+ y 2 1)a 2 n+(x 2 2+ y 2 2)b 2 n+2(x1x2+ y1y2)anbn = (x1an+x2bn) 2+(y1an+ y2bn) 2. Ïd, (Ø�y. 5
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