4題等比問題..數學高手請進!!!(20分)－bgdwze3的部落格

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4題等比問題..數學高手請進!!!(20分)

發問:

1.在小於100的正整數中..有多少個不能被7整除?? 2同3:http://img267.imageshack.us/i/maths6003.jpg/ 4:http://img262.imageshack.us/i/maths7002.jpg/ 姐係29題..30唔洗..小弟我完全唔識做..請各位數學高手救下我..多謝...

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(1) 100/7 = 14餘2 小於100的7的倍有14 個其餘86個不能被7整除 (27) (a)(i) Q1 = P(1 + r%)(1/3) = (P/3)(1 + r%); 餘下(2P/3)(1 + r%) Q2 = (2P/3)(1 + r%)(1 + r%)(1/3) = (2P/9)(1 + r%)^2; 餘下(4P/9)(1 + r%)^2 Q3 = (4P/9)(1 + r%)^2(1 + r%)(1/3) = 4P/27(1 + r%)^3 (ii) 公比 = [(2P/9)(1 + r%)^2] / [(P/3)(1 + r%)] = (2/3)(1 + r%) (b) (i) Q3 = 4P/27(1 + r%)^3 = 27P/128 (1 + r%)^3 = 27*27/512 1 + r% = 9/8 r% = 12.5% r = 12.5 (ii) 公比 = (2/3)(9/8) = 0.75 Q1 + ... + Q10 = (20000/3)(1.125)*[1 - 0.75^10]/[1 - 0.75] = $28311 (iii) Q1 + Q2 + ... = (20000/3)(1.125)/(1 - 0.75) = $30000 (28) (a)(i) BA1 / A1B1 = tanθ (BC - A1C)/A1B1 = tanθ (a - A1B1)/A1B1 = 1/3 3a - 3A1B1 = A1B1 4A1B1 = 3a A1B1 = 3a/4 (ii) A2B2 = (3/4)A1B1 = 9a/16 (b) R1 = a^2 R2 = (3a/4)^2 = 9a^2/16 R3 = (9a/16)^2 = 81a^3/256 R1*R3 = 81a^4/256 = (9a^2/16)^2 = R2^2 所以R1, R2, R3成一等比數列。公比 = (9a^2/16) / a^2 = 9/16 (ii) R1 + R2 + ... + Rn = R1[1 - (9/16)^n]/(1 - 9/16) = 16a^2[1 - (9/16)^n] / 7 (iii) 無限項之和 = 16a^2/7 (30) (a) PQ = 3, QR = 5, PR = 4 因 PQ^2 + PR^2 = QR^2 PQR為直角三角形，面積 = (1/2)(3)(4) = 6cm^2 PM1N1 及PQR為相似三角形，邊長比例 = 1 : 2 面積比例 = 1: 4 PM1N1 : PQR = 1 : 4 A1 : PQR = (4 - 1) : 4 = 3 : 4 A1面積 = 6cm^2 * 3 / 4 = 4.5 cm^2 (b) PM1N1面積 = 6 cm^2 /4 = 1.5 cm^2 以和(a)相同的方法，A2面積 = PM1N1面積 *3/4 = 1.125 cm^2 (c) 公比為 1.125 / 4.5 = 1/4 (c) A1 + A2 + A3 + ... = A1[1 / (1 - 1/4)] = 4.5 * 4/3 = 6 cm^2

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