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數學知識交流---求值
求 1 + 11 + 111 + ... + 111...11 [ (n - 1) 個 1 ] + 111...11 [ n 個 1 ] 若 (1) n = 34 (2) n = 139 (3) n = x - 1 ， 請詳細說明答案。
最佳解答:
以等比序列公式求: 1 = (101 - 1)/(10 - 1) = (101 - 1)/9 11 = 1 + 10 = (102 - 1)/(10 - 1) = (102 - 1)/9 111 = 1 + 10 + 102 = (103 - 1)/(10 - 1) = (103 - 1)/9 . . . 11...1 (n 個 1) = 1 + 10 + 102 + 103 + ... + 10n-1 = (10n - 1)/9 所以: 1 + 11 + 111 + ... + 11...1 = (1/9) [(101 - 1) + (102 - 1) + ... + (10n - 1)] = (1/9) [(101 + 102 + ... + 10n) - (1 - 1 - ... - 1)] = (1/9) [10 x (1 + 101 + ... + 10n-1) - n] = (1/9) [10 x (10n - 1)/9 - n] = 10 x (10n - 1)/81 - n/9 所以: (1) 代 n = 34, 總和 = 10 x (1034 - 1)/81 - 34/9 (2) 代 n = 139, 總和 = 10 x (10139 - 1)/81 - 139/9 (3) 代 n = x - 1, 總和 = 10 x (10x-1 - 1)/81 - (x - 1)/9
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數學知識交流---求值
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發問:求 1 + 11 + 111 + ... + 111...11 [ (n - 1) 個 1 ] + 111...11 [ n 個 1 ] 若 (1) n = 34 (2) n = 139 (3) n = x - 1 ， 請詳細說明答案。
最佳解答:
以等比序列公式求: 1 = (101 - 1)/(10 - 1) = (101 - 1)/9 11 = 1 + 10 = (102 - 1)/(10 - 1) = (102 - 1)/9 111 = 1 + 10 + 102 = (103 - 1)/(10 - 1) = (103 - 1)/9 . . . 11...1 (n 個 1) = 1 + 10 + 102 + 103 + ... + 10n-1 = (10n - 1)/9 所以: 1 + 11 + 111 + ... + 11...1 = (1/9) [(101 - 1) + (102 - 1) + ... + (10n - 1)] = (1/9) [(101 + 102 + ... + 10n) - (1 - 1 - ... - 1)] = (1/9) [10 x (1 + 101 + ... + 10n-1) - n] = (1/9) [10 x (10n - 1)/9 - n] = 10 x (10n - 1)/81 - n/9 所以: (1) 代 n = 34, 總和 = 10 x (1034 - 1)/81 - 34/9 (2) 代 n = 139, 總和 = 10 x (10139 - 1)/81 - 139/9 (3) 代 n = x - 1, 總和 = 10 x (10x-1 - 1)/81 - (x - 1)/9
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