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1. A sequence consists of 10 terms, all of which are +ve integers. The 1st term is p and the 2nd term is q, with q greater than p. Each term thereafter is the sum of the 2 terms immediately preceding it and the 7th term is 181. Determine the sum of the terms.2. Solve (x^2+3x+2)(x^2+7x+12)+(x^2+5x-6)=0Thanks! 顯示更多 1. A sequence consists of 10 terms, all of which are +ve integers. The 1st term is p and the 2nd term is q, with q greater than p. Each term thereafter is the sum of the 2 terms immediately preceding it and the 7th term is 181. Determine the sum of the terms. 2. Solve (x^2+3x+2)(x^2+7x+12)+(x^2+5x-6)=0 Thanks! 更新: 3. The first two terms in a sequence are 1 and 3. If any 2 consecutive terms of this sequence are a and b then the next term is b+1/a. a) What is the 100th term? b) Determine the sum of the 1st 152 terms. Thx! 更新 2: Given the equation ax^2+bx+c=0. If one root is 5 times the other, which of the following must be true? a) 5b^2=36ac b) 36b^2=5ac c) 5b^2=6ac d) 25b^2=36ac e) 3b^2=25ac
(1) Simple calculation shows that the ten terms are: p, q, p+q, p+2q, 2p+3q, 3p+5q, 5p+8q (= 181), 8p+13q, 13p+21q, 21p+34q The sum of these 10 terms are 55p+88q.【Check by yourself】 Note that we are lucky enough that the sum = 55p+88q = 11 (5p+8q) = 11 x 7th term = 11 x 181 = 1991 (2) The way I do this question is try to "make" x2+5x-6 appear in the first term. (x2+3x+2)(x2+7x+12)+(x2+5x-6)=0 (x2+5x-6-2x+8)(x2+5x-6+2x+18)+(x2+5x-6)=0 [(x2+5x-6)+(-2x+8)][(x2+5x-6)+(2x+18)]+(x2+5x-6)=0 (x2+5x-6)2 + (2x+18)(x2+5x-6) + (-2x+8)(x2+5x-6) + (-2x+8)(2x+18) + (x2+5x-6) = 0 (x2+5x-6)2 + (2x+18-2x+8)(x2+5x-6) + (-4x2-36x+16x+144) + (x2+5x-6) = 0 (x2+5x-6)2 + 26(x2+5x-6) + (-3x2-15x+138) = 0 (x2+5x-6)2 + 26(x2+5x-6) + (-3)(x2+5x-6) + 120 = 0 (x2+5x-6)2 + 23(x2+5x-6) + 120 = 0 [(x2+5x-6)+15][(x2+5x-6)+8] = 0 (x2+5x+9)(x2+5x+2) = 0 x2+5x+9=0 or x2+5x+2=0 x={-5±√[52-4(1)(9)]}/2 or x={-5±√[52-4(1)(2)]}/2 x=(-5±√11 i)/2 or x=(-5±√17)/2 where i=√(-1) is the imaginary unit (3) I think there's some bad notation in you question. The recursive definition should be "(b+1)/a" but not "b+1/a" (a) Let an denotes the n-th term. By listing the first few terms, we have, a1 = 1 a2 = 3 a3 = (3+1)/1 = 4 a4 = (4+1)/3 = 5/3 a5 = (5/3 + 1) / 4 = (8/3) / 4 = 2/3 a6 = (2/3 + 1) / (5/3) = (5/3) / (5/3) = 1 a7 = (1+1) / (2/3) = 2 / (2/3) = 3 So the terms are periodic with a1 = a6 = a11 = a16 = ... Since 100 = 19 x 5 + 5, a100 = a5 = 2/3 (b) The sum from a1 to a5 = 1 + 3 + 4 + 5/3 + 2/3 = 31/3 In the first 152 terms, there are totally 30 sets of (a1 to a5) So the sum of first 152 terms = 30 x 31/3 + a151 + a152 = 310 + a1 + a2 = 310 + 1 + 3 = 314 Hope it helps! ^^ 2006-12-28 11:59:46 補充： 4.Let α and 5α be the roots.Then α 5α = -b/aα = -b/(6a) ---(1)and α(5α) = c/a5α2 = c/a ---(2)Sub. (1) into (2),5[-b/(6a)]2 = c/a5b2 = c/a * 36a25b2 = 36acThe answer is (a).
其他解答:
付詳情解題63D0B758E2D502CC
Challenging!
發問:
1. A sequence consists of 10 terms, all of which are +ve integers. The 1st term is p and the 2nd term is q, with q greater than p. Each term thereafter is the sum of the 2 terms immediately preceding it and the 7th term is 181. Determine the sum of the terms.2. Solve (x^2+3x+2)(x^2+7x+12)+(x^2+5x-6)=0Thanks! 顯示更多 1. A sequence consists of 10 terms, all of which are +ve integers. The 1st term is p and the 2nd term is q, with q greater than p. Each term thereafter is the sum of the 2 terms immediately preceding it and the 7th term is 181. Determine the sum of the terms. 2. Solve (x^2+3x+2)(x^2+7x+12)+(x^2+5x-6)=0 Thanks! 更新: 3. The first two terms in a sequence are 1 and 3. If any 2 consecutive terms of this sequence are a and b then the next term is b+1/a. a) What is the 100th term? b) Determine the sum of the 1st 152 terms. Thx! 更新 2: Given the equation ax^2+bx+c=0. If one root is 5 times the other, which of the following must be true? a) 5b^2=36ac b) 36b^2=5ac c) 5b^2=6ac d) 25b^2=36ac e) 3b^2=25ac
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最佳解答:(1) Simple calculation shows that the ten terms are: p, q, p+q, p+2q, 2p+3q, 3p+5q, 5p+8q (= 181), 8p+13q, 13p+21q, 21p+34q The sum of these 10 terms are 55p+88q.【Check by yourself】 Note that we are lucky enough that the sum = 55p+88q = 11 (5p+8q) = 11 x 7th term = 11 x 181 = 1991 (2) The way I do this question is try to "make" x2+5x-6 appear in the first term. (x2+3x+2)(x2+7x+12)+(x2+5x-6)=0 (x2+5x-6-2x+8)(x2+5x-6+2x+18)+(x2+5x-6)=0 [(x2+5x-6)+(-2x+8)][(x2+5x-6)+(2x+18)]+(x2+5x-6)=0 (x2+5x-6)2 + (2x+18)(x2+5x-6) + (-2x+8)(x2+5x-6) + (-2x+8)(2x+18) + (x2+5x-6) = 0 (x2+5x-6)2 + (2x+18-2x+8)(x2+5x-6) + (-4x2-36x+16x+144) + (x2+5x-6) = 0 (x2+5x-6)2 + 26(x2+5x-6) + (-3x2-15x+138) = 0 (x2+5x-6)2 + 26(x2+5x-6) + (-3)(x2+5x-6) + 120 = 0 (x2+5x-6)2 + 23(x2+5x-6) + 120 = 0 [(x2+5x-6)+15][(x2+5x-6)+8] = 0 (x2+5x+9)(x2+5x+2) = 0 x2+5x+9=0 or x2+5x+2=0 x={-5±√[52-4(1)(9)]}/2 or x={-5±√[52-4(1)(2)]}/2 x=(-5±√11 i)/2 or x=(-5±√17)/2 where i=√(-1) is the imaginary unit (3) I think there's some bad notation in you question. The recursive definition should be "(b+1)/a" but not "b+1/a" (a) Let an denotes the n-th term. By listing the first few terms, we have, a1 = 1 a2 = 3 a3 = (3+1)/1 = 4 a4 = (4+1)/3 = 5/3 a5 = (5/3 + 1) / 4 = (8/3) / 4 = 2/3 a6 = (2/3 + 1) / (5/3) = (5/3) / (5/3) = 1 a7 = (1+1) / (2/3) = 2 / (2/3) = 3 So the terms are periodic with a1 = a6 = a11 = a16 = ... Since 100 = 19 x 5 + 5, a100 = a5 = 2/3 (b) The sum from a1 to a5 = 1 + 3 + 4 + 5/3 + 2/3 = 31/3 In the first 152 terms, there are totally 30 sets of (a1 to a5) So the sum of first 152 terms = 30 x 31/3 + a151 + a152 = 310 + a1 + a2 = 310 + 1 + 3 = 314 Hope it helps! ^^ 2006-12-28 11:59:46 補充： 4.Let α and 5α be the roots.Then α 5α = -b/aα = -b/(6a) ---(1)and α(5α) = c/a5α2 = c/a ---(2)Sub. (1) into (2),5[-b/(6a)]2 = c/a5b2 = c/a * 36a25b2 = 36acThe answer is (a).
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付詳情解題63D0B758E2D502CC
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