answer:(1) Let b_n = frac {1}{a_n - frac{1}{2}}. Thus, we have a_n = frac {1}{b_n} + frac {1}{2}. Since (1-a_n)a_{n+1}= frac {1}{4}, then: (1- left( frac {1}{b_n} + frac {1}{2} right)) left( frac {1}{b_{n+1}} + frac {1}{2} right) = frac {1}{4}. This simplifies to: b_{n+1} - b_n = -2. Given that b_1 = frac {1}{ frac{1}{4} - frac{1}{2}} = -4, we can deduce that {b_n} is an arithmetic sequence with a first term of -4 and a common difference of -2. Hence, the sequence left{ frac {1}{a_n - frac{1}{2}} right} is an arithmetic sequence. (2) Knowing from (1) that b_n = -4 - 2(n-1) = -2n - 2, we get a_n = frac {1}{b_n} + frac {1}{2} = - frac {1}{2n + 2} + frac {1}{2} = frac {n}{2(n+1)}. Therefore: frac {a_{n+1}}{a_n} = frac {n+1}{2(n+2)} cdot frac {2(n+1)}{n} = frac {(n+1)^2}{n(n+2)} = 1+ frac {1}{n(n+2)} = 1+ frac {1}{2}left( frac {1}{n} - frac {1}{n+2} right). Summing this from 1 to n, we find: frac {a_2}{a_1} + frac {a_3}{a_2} + ldots + frac {a_{n+1}}{a_n} = n+ frac {1}{2}left(1 - frac {1}{3} + frac {1}{2} - frac {1}{4} + ldots + frac {1}{n} - frac {1}{n+2}right). The series telescopes to: n+ frac {1}{2}left(1+ frac {1}{2} - frac {1}{n+1} - frac {1}{n+2}right), which is less than n+ frac {3}{4}. Hence we have boxed{n+ frac {1}{2}left(1+ frac {1}{2} - frac {1}{n+1} - frac {1}{n+2}right) < n+ frac {3}{4}}.

answer:To solve for left(-7right)^{4}, we need to multiply -7 by itself four times. This is represented as: [ left(-7right)^{4} = (-7) times (-7) times (-7) times (-7) ] Looking at the options provided: A: (-7)times 4 is incorrect because it represents -7 multiplied by 4, not -7 multiplied by itself four times. B: -7times 7times 7times 7 is incorrect because it does not maintain the negative sign across all multiplications. C: (--7)+left(-7right)+left(-7right)+left(-7right) is incorrect because it represents addition of -7 four times, not multiplication. D: (-7)times left(-7right)times left(-7right)times left(-7right) correctly represents -7 multiplied by itself four times. Therefore, the correct expression that represents left(-7right)^{4} is: [ boxed{D} ]

answer:According to the problem, we have begin{align*} 3a + b &= 8, -2a + b &= 3. end{align*} Solving these equations, we get begin{align*} a &= 1, b &= 5. end{align*} Therefore, the equation of the linear function is y = x + 5. Since the graph of the linear function passes through point C(-3, c), -3 + 5 = c, we find c = 2. Therefore, a^2 + b^2 + c^2 - ab - bc - ac = 1^2 + 5^2 + 2^2 - 1 times 5 - 5 times 2 - 1 times 2 = 1 + 25 + 4 - 5 - 10 - 2 = 30 - 17 = 13. Hence, the answer is boxed{13}.

answer:Let's analyze each option step by step according to the given rules and the solution provided. # Option A: If a > b and c > d, then a+c > b+d - Given: a > b and c > d - By the additive property of inequalities, if you add the same quantity to both sides of an inequality, the direction of the inequality does not change. - Therefore, adding c to a > b gives a+c > b+c - Similarly, adding a to c > d gives a+c > a+d - Combining these, we have a+c > b+d - Hence, option A is boxed{text{correct}}. # Option B: If a > b and c > d, then ac > bd - Given: a > b and c > d - Counterexample: Let a = 0 > -1 = b and c = 0 > -1 = d - Then, ac = 0cdot0 = 0 and bd = (-1)cdot(-1) = 1 - We find that 0 < 1 which contradicts the statement that ac > bd - Hence, option B is boxed{text{incorrect}}. # Option C: If a < b, then ac^2 < bc^2 - Given: a < b - Counterexample: Let c = 0 - Then, ac^2 = acdot0^2 = 0 and bc^2 = bcdot0^2 = 0 - We find that ac^2 = bc^2 which contradicts the statement that ac^2 < bc^2 - Hence, option C is boxed{text{incorrect}}. # Option D: If a > b > 0 and c < 0, then frac{c}{a} > frac{c}{b} - Given: a > b > 0 and c < 0 - Since a > b > 0, we have 0 < frac{1}{a} < frac{1}{b} - Multiplying an inequality by a negative number reverses the direction of the inequality. - Multiplying by c < 0, we get ccdotfrac{1}{a} > ccdotfrac{1}{b} - Which simplifies to frac{c}{a} > frac{c}{b} - Hence, option D is boxed{text{correct}}. Therefore, the correct options are boxed{text{A and D}}.