a r 1 qa m r a m n a m a n 1 a m N u � n m ta s ử d ng ụ thu t ậ toán m nq r v

A r 1 qa m r a m n a m a n 1 a m n u ? n m ta s ử

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a r + 1 qa m + r a m = n a m a n + 1 a m −→ . N u ế n < m , ta s d ng thu t toán m = nq + r v i r < m , ta có r + n < m + n , ta l i s d ng gi thi t quy ế n p: a r a n + 1 a r = 1 2 n > 1 + 2 n = 1 2 n = 1 . 2 n 1 . 2 2 V i n 2 có n 2 2 2 , nên 2 n 1 1 và 1 2 n > 1 . 2 n 1 1 2 = 2 . 2 2 n 2 2 2 n + 1 > 0 −→ 1 2 2 n . 2 n > 0 −→ 1 > 2 2 n 2 n −→ 2 > 2 2 n . B t đ ng th c cu i cùng luôn đúng. V y bài toán đã đ c ượ n n r m + r m m n m
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r n −→ na m = na qn + r n ( qa n + a r + q ) ≤ nqa n + nq + nr r nqa n + nq + nr a n + 1 n a m = nqa n + nq + ra n + r = m ( a n + 1 ) −→ m minh. Bài toán 19(Shortlist 1989 [5]) a n + 1 . V y bài toán đã đ c ch ng ượ n Cho t p các s th c { a 0 , a 1 , · · ·, a n } th a mãn các đi u ki n sau: (i) a 0 = a n = 0, (ii) V i 1 k n 1, Ch ng minh r ng : c 1 . 4 n n 1 a k = c + a i k .( a i + a i + 1 ) i = k L i gi i k Đ t Sk = i ai ( k = 0 , 1 , · · ·, n ) . Khi đó: S = n 1 a = nc + n 1 n 1 a =
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( a + a ) n k i k i i + 1 k = 0 k = 0 i = k = nc + n−1 i a ( a + a ) = nc + n−1 a + a ) i a i i k i i + 1 i i + 1 i k = 0 k = 0 i = 0 k = 0 = nc + n−1 a + a i ) a , v i t = i k i i = 0 i + 1 i t = 0 = nc + n−1 a + a ) S i i = 0 i + 1 1 = nc + ( S 1 S 0 + ( S 2 S 0 ) S 1 + ( S 3 S 1 ) S 2 + · · · + ( S n S n 2 ) S n 1 ) = nc + S 2 (vì S n 1 = S n ). Do v y ta có : S 2 S n + nc = 0 S n là s th c nên: 1 4 nc . Ta có đi u ph i ch ng minh. Bài toán 20( Shortlist 1994 [5]) a 2 Cho a 0 = 1994 và a n + 1 = 0 n 998 . n a n + 1 a 2 v i n 0. Ch ng minh r ng | a n ∫ = 1994 n v i L i gi i 1 Ta có a n a n + 1 = a n n a n + 1 = 1 a n + 1 > 0. Do v y, a 0 > a 1 > · · · > a n > · · · L i có : a n = a 0 +( a 1 a 0 )+···+( a n a n 1 1 ) = 1994 n + 1 +···+ > 1994 n . a 0 + 1 a n 1 + 1 V i 1 n 998, ta có : 1 1 n 998 998 a 0 + 1 + · · · + a n < 1 + 1 a n < < = 1 1 + 1 a 997 + 1 1994 997 + 1 Do v y | a n ∫ = 1994 n . ( ( ( n n
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Bài toán 21(Shortlist 1996)
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Cho a > 2. Ta đ nh nghĩa nh sau: ư a 1 a 1 a . a 2 2 Σ a 0 = , 1 = , n + 1 = n a 2 n Ch ng minh r ng v i m i k ∈ N ta có 1 1 a 0 + · · · + a k T a > 2 ta th vi t ế a = b + 1 b 1 . 2 + a a 2 4 Σ L i gi i v i b là m t s th c d ng. Khi đó ta có : ươ a 2 2 = b 2 + 1 và: a 2 = ( a 2 2 ) a = . b 2 + 1 Σ . b + 1 Σ . . a 2 Σ 2 Σ . . 2 1 Σ 2 Σ = . b 4 + 1 Σ . b 2 + 1 Σ . b + 1 Σ . b 4 Ti p t c quá trình trên ta thu ế đ c: ượ b 2 b a = . b 2 n−1 + 1 Σ · · · . b 2 + 1 Σ . b + 1 Σ . n Do v y, b 2 n 1 b 2 b n 1 b b 3 b 2 n 1 a i = 1 + b 2 + 1 + ( b 2 + 1 )( b 4 + 1 ) + · · · + ( b 2 + 1 )( b 4 + 1 ) · · · ( b 2 n + 1 ) Ta l i có : 1 ( a + 2 a 2 4 ) = 1 . b + 1 + 2 . b 1 ΣΣ = 1 1 Do v y ta ph i ch ng minh v i m i b > 0 thì, b 2 b 4 n 1 2 b 2 b 2 b a 3 = 2 a 2 = b + b 2 2 a 2 a 1 a 1 i = 2 2 b b + b .
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b 2 n 1 + b 2 + ( 1 + b 2 )( 1 + b 4 ) + · · · + ( 1 + b 2 )( 1 + b 4 ) · · · ( 1 + b 2 n ) < 1 .
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Mà v i m i s th c d ng ươ a 1 , a 2 , · · ·, a n , n a j 1 Do v y : ( 1 + a 1 ) · · · ( 1 + a j ) = 1 ( 1 + a 1 )( 1 + a
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