1 k 1 n 2 và a 1 k 1 n 2 k 3 n 1 do v y ậ ta ch

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1 = ( k + 1 ) n 2 , và a 1 =
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( k + 1 ) n 2 ( k + 3 n ) − 1 . Do v y, ta ch c n l y k sao cho: ( k + 1 ) n 2 ( k + 3 n ) − 1 k n 1 > 0 . Nhìn v ế trái c a đa th c n k , h s c a k n 1 b ng 0 nh ng h s c a ư k n 2 b ng 1. Có nghĩa là bi u th c d ng ươ v i k đ l n ủ ớ và ta có th tìm đ c ượ dãy a 1 , a 2 , · · ·, a n b 1 , b 2 , · · ·, b n th a mãn yêu c u bài toán. V i n = 5, ta tìm k sao cho: ( k + 1 ) 3 ( k 2 ) − 1 k 4 > 0 . Ta th y r ng k = 5 th a mãn và ta có: 625 < 647 < 750 < 863 < 900 < 1079 < 1080 < 1295 < 1296 < 1511 . Bài toán 12(Romania 2000 [10]) : Cho a là m t s th c d ng ươ { x n } ( n 1 ) là m t dãy s th c sao cho x 1 = a n 1 x n + 1 ( n + 2 ) x n kx k , k = 1 v i m i n 1. Ch ng minh r ng t n t i s nguyên d ng ươ n sao cho x n > 1999!. L i gi i Ta s ch ng minh b ng quy n p theo n 1 r ng: n x n + 1 > kx k > a . n ! k = 1 V i n = 1, ta có x 2 3 x 1 > x 1 = a . Gi s r ng kh ng đ nh đúng đ n ả ử ằ ế n . Khi đó: n n x n + 2 ≥ ( n + 3 ) x n + 1 k = 1 kx k = ( n + 1 ) x n + 1 + 2 x n + 1 k = 1 kx k > ( n + 1 ) x n + 1 + 2 n k = 1 kx k n k = 1 kx k = n + 1 k = 1 kx k , H n n a, ơ x 1 > 0 theo đ nh nghĩa x 2 , x 3 , · · ·, x n cũng d ng theo gi thi t quy n p; ươ ế do v y x n + 2 > ( n + 1 ) x n + 1 > ( n + 1 )( a . n ! ) = a .( n + 1 ) !. Nh ư v y kh ng đ nh đ c ch ng minh hoàn ượ toàn. V i n đ l n, ta có ủ ớ x n + 1 > n ! . a > 1999!.
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Bài toán 13(APMO1999) . Cho a 1 , a 2 , · · · là dãy các s th c th a mãn: a i + j a i + a j v i m i i , j = 1 , 2 , · · · . Ch ng minh r ng: a 1 + a 2 + a 3 + · · · + a n a n . 2 3 n v i m i s nguyên d ng ươ n . Ta ch ng minh b đ sau: L i gi i B đ : N u ế m , n là các s nguyên d ng ươ v i m n , khi đó a 1 + a 2 + + a n n ( n + 1 ) 2 m . a m . Th t v y, ta s ch ng minh k t qu cho ế m = n b ng cách c ng các b t đ ng th c : a 1 + a n 1 a n , a 2 + a n 2 a n , · · ·, a n 1 + a 1 a n , 2 a n 2 a n , và chia cho 2. V i s nguyên d ng ươ j , vi t ế j = · · · . Khi đó b t đ ng th c v i m = 1 + 2 + · · a · + j n = j = k + 1 là t ng đ ng ươ ươ v i j j k k + 1 ; nên khi m n ta có n n + a m 1 ≥ · · · ≥ m m . Nh v y b đ đ c ch ng minh. ư ượ T b t đ ng th c đã ch ng minh ta bi u di n a 1 + a 2 + · · · + a n nh t ng Abel và ư ổ 2 n sau đó áp d ng b đ nhi u l n: a 2 a n 1 n 1 . 1 1 Σ 1 n ( n + 1 ) n 1 1 j ( j + 1 ) n . a n + j = 1 j ( j + 1 ) . 2 n a n = a n , · · · a 1 + a 2 + + a j n = n ( a 1 + a 2 + · · · + a n ) + j ( a 1 + a 2 + · · · + a j ) a 1 + 2 + · · · + j = j + j = j + 2
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Nh v y bài toán đã đ c ch ng minh. ư ượ Bài toán 14(Canada 2000 [8]) .
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