Bài toán 1 dãy s ố h n đ c cho b i đi u ki n

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Bài toán 1. Dãy s ( h n ) đ c cho b i đi u ki n: ượ h 1 = 1 h n = 2 n 2 2 n 1 Đ t S n = hi ; n ∈ N . Hãy ch ng minh r ng:lim Sn < 1 , 03 2 n n = t . 1 1 h ; + n
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L i gi i i = 1 n Ta có: h 1 = 1 = sin = sin h 2 = sin Ta s ch ng minh r ng: h n = sin . 1 . 1 sin . 1 cos Gi s r ng:sin h k = sin 3 . 2 h k + 1 = , 3 . 2 2 = 3 . 2 k 2 = sin 3.2 n M t khác:sin x < x ; x . 0; Σ . Nên: S n 1 = h = + sin + ... + sin < 2 1 1 n i i = 1 2 3 . 2 n 3 . 2 n 2 + 3 . 2 2 + ... + 3 . 2 n < 2 + 3 . 2 Do Sn là dãy tăng nên lim S n ≤ + < 1 , 03 đpcm. n 2 3 . 2 Bài toán 2 Cho dãy ( u n ) đ nh b i:  u 1 = 3 Tính u 2003 n + 1 1 + . 1 2 Σ u n Nh n xét : v i gi thi t c a bài ta liên t ng ế ưở ngay đ n ế công th c: tan ( a + b ) = tan a + tan b . 1 tan a tan b Đ ng th i ta còn có 2 1 = tan 8 , u 1 = tan 3 L i gi i  u 1 = 3 1 + 1 2 u n 2 tg u + tg 1 tg tg . 2 Σ 8 tg 2 1 T (*) ta có: u 8 1 2 6 3 . 2 3 . 2 3 . n 2 k , k 1 u u n + = 2 1 Ta có: u n + 1 = 2 1 (∗) Ta đã bi t: ế tg 8 = 2 1 . Σ u + n + 1 ( ) n = 4 = . 8 = 1 2 8 = Theo nguyên lý quy n p, t (1) và u 1 = 3.suy ra Suy ra: u n = tg Σ + ( n 1 ) Σ
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tg 8 1 u n tg 8 V y: u 2003 = tg . + 2002 Σ = tg . + Σ = − . 2 + 3 Σ Tìm lim u n n Cho dãy n xác đ nh b i: u n = 2 n . 2 . 2 + ... 2 Bài gi i. Bài toán 3. u 8 3 4 3 8 Bài toán 3. 3
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Đây là bài toán đ n gi n và quen thu c. Ta s ch ng minh: ơ v n = . 2 . 2 + ... 2 = 2 cos ( 1 ) . Rõ ràng v i n = 1 thì (1) hi n nhiên đúng. Gi s đúng khi n = k, nghĩa là: ả ử v 2 cos . 2 k + 1 Xét: v k + 1 = 2 + v k = . 2 + 2 cos 2 2cos 2 2 k + 2 = 2 cos V y (1) đúng khi n = k+1, suy ra (1) đúng v i m i n. Ta có: u n = 2 n . 2 2 + ... 2 = 2 n + 1 . sin = 1 . 2 n + 2 . sin n n 2 2 n + 2 1 sin = lim n 2 2 n + 2 2 n + 2 lim u = n 2 Bài toán 4. Cho dãy s xác đ nh b i: a 0 = 1; a 1000 = 0 a n + 1 = 2 a 1 . a n a n 1 Tính: a 1999 + a 1 L i gi i. + N u ế thay n=2 thì ta đ c ượ a 2 = 2 a 2 1,v y nên n u ế mu n s d ng l ng ượ giác( đây là hàm cos, vì cos 2 a = 2 cos 2 a 1 )) ta c n ph i ch ng minh đ c ta ượ c n ph i ch ng minh đ c ượ | a 1 | ≤ 1 thì m i có th đ t a 1 = cosa . 1 suy ra | a 3 | = | 2 a 1 a 2 a 1 | > | 2 a 2 1 | > 1,..., | a 1000 | > 1 (trái v i gi thi t). ế V y nên | a 1 | ≤ 1, đ t a 1 = cosa . B ng quy n p, ta ch ng minh đ c r ng: ượ a n + 1 = c os ( n + 1 ) a . Ta có: a 1000 = c os1000 a = 0 1000 a = + k 2 2 k = . = .
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