answer:As r^7 = 1, r is a seventh root of unity but not 1, thus r^7 - 1 = 0. This factors as: [(r - 1)(r^6 + r^5 + r^4 + r^3 + r^2 + r + 1) = 0.] Given r neq 1, it must be that r^6 + r^5 + r^4 + r^3 + r^2 + r + 1 = 0. To compute the product, pair the factors accordingly: [ (r - 1)(r^2 - 1)(r^3 - 1)(r^4 - 1)(r^5 - 1)(r^6 - 1) = [(r - 1)(r^6 - 1)][(r^2 - 1)(r^5 - 1)][(r^3 - 1)(r^4 - 1)]. ] Expanding these, since r^7 = 1, we can replace r^7 with 1: [ [(r^7 - r - r^6 + 1)][(r^7 - r^2 - r^5 + 1)][(r^7 - r^3 - r^4 + 1)]. ] Simplifying, each becomes (1 - r - r^6 + 1)(1 - r^2 - r^5 + 1)(1 - r^3 - r^4 + 1): [ (2 - r - r^6)(2 - r^2 - r^5)(2 - r^3 - r^4). ] When expanded, all terms involving r to any power sum up because we know that r^6 + r^5 + r^4 + r^3 + r^2 + r + 1 = 0. [ 4 - 2(r + r^2 + r^3 + r^4 + r^5 + r^6) + (r + r^2 + r^3 + r^4 + r^5 + r^6)^2 = 7 - (r + r^2 + r^3 + r^4 + r^5 + r^6) = boxed{7}. ]

answer:Given frac{a}{b}=2, we aim to find the value of frac{a}{a-b}. Starting from the given ratio: frac{a}{b} = 2 This implies that: a = 2b Substituting a = 2b into frac{a}{a-b}, we get: frac{a}{a-b} = frac{2b}{2b-b} Simplifying the denominator: = frac{2b}{b} This simplifies to: = 2 Therefore, the value of frac{a}{a-b} is boxed{2}.

answer:To determine the range of real number t for which the function f(x)=sinleft(2x+frac{pi}{6}right) is monotonically increasing on the interval left[-t,tright], we start by considering the general condition for a sine function to be monotonically increasing. For a sine function, this occurs within one period where the argument of the sine function increases from -frac{pi}{2} to frac{pi}{2}. Given f(x)=sinleft(2x+frac{pi}{6}right), we set up the inequality to find the interval where this function is increasing: [ 2kpi -frac{pi}{2}leqslant 2x+frac{pi}{6}leqslant 2kpi +frac{pi}{2}, quad kin mathbb{Z}. ] Solving this inequality for x gives us: [ kpi -frac{pi}{3}leqslant xleqslant kpi +frac{pi}{6}, quad kin mathbb{Z}. ] This means the function f(x) is monotonically increasing in intervals of the form [kpi -frac{pi}{3}, kpi +frac{pi}{6}] for any integer k. Given that f(x) must be monotonically increasing on the interval left[-t,tright], this interval must be a subset of one of the intervals where f(x) is increasing, specifically [kpi -frac{pi}{3}, kpi +frac{pi}{6}] for some kin mathbb{Z}. Considering the case when k=0, we have the interval [-frac{pi}{3},frac{pi}{6}]. For left[-t,tright] to be a subset of this interval, t must satisfy: [ 0 < t leqslant frac{pi}{6}. ] Therefore, the range of real number t for which the function f(x) is monotonically increasing on the interval left[-t,tright] is boxed{text{D: }left(0,frac{pi}{6}right]}.

answer:1. Given the sequence (langle x_n rangle) defined by: [ x_1 = 1, quad x_n = dfrac{2n}{(n-1)^2} sum_{i=1}^{n-1} x_i ] we need to show that the sequence (y_n = x_{n+1} - x_n) has finite limits as (n to infty). 2. First, we express the sum (sum_{i=1}^{n-1} x_i) in terms of (x_n): [ sum_{i=1}^{n-1} x_i = frac{(n-1)^2 x_n}{2n} ] This follows directly from the given recurrence relation. 3. Next, we express (sum_{i=1}^{n} x_i) in terms of (x_{n+1}): [ sum_{i=1}^{n} x_i = frac{n^2 x_{n+1}}{2(n+1)} ] 4. Subtracting the first expression from the second, we get: [ x_n = frac{n^2 x_{n+1}}{2(n+1)} - frac{(n-1)^2 x_n}{2n} ] 5. Simplifying the above equation, we find: [ x_n = frac{n^2 x_{n+1}}{2(n+1)} - frac{(n-1)^2 x_n}{2n} ] [ 2n(n+1)x_n = n^2 x_{n+1} - (n-1)^2 x_n ] [ 2n(n+1)x_n + (n-1)^2 x_n = n^2 x_{n+1} ] [ (2n(n+1) + (n-1)^2)x_n = n^2 x_{n+1} ] [ (2n^2 + 2n + n^2 - 2n + 1)x_n = n^2 x_{n+1} ] [ (3n^2 + 1)x_n = n^2 x_{n+1} ] [ x_{n+1} = frac{3n^2 + 1}{n^2} x_n ] [ x_{n+1} = left(3 + frac{1}{n^2}right) x_n ] 6. Now, we express (y_n = x_{n+1} - x_n): [ y_n = x_{n+1} - x_n = left(3 + frac{1}{n^2}right)x_n - x_n ] [ y_n = left(3 + frac{1}{n^2} - 1right)x_n ] [ y_n = left(2 + frac{1}{n^2}right)x_n ] 7. As (n to infty), (frac{1}{n^2} to 0), so: [ y_n approx 2 x_n ] 8. To show that (y_n) has a finite limit, we need to show that (x_n) grows at a controlled rate. From the recurrence relation: [ x_{n+1} = left(3 + frac{1}{n^2}right)x_n ] we see that (x_n) grows exponentially, but the factor (left(3 + frac{1}{n^2}right)) approaches 3 as (n to infty). 9. Therefore, (y_n = left(2 + frac{1}{n^2}right)x_n) will also grow exponentially, but the growth rate is controlled by the factor (2 + frac{1}{n^2}), which approaches 2. 10. Finally, we conclude that (y_n) has a finite limit as (n to infty): [ lim_{n to infty} y_n = lim_{n to infty} left(2 + frac{1}{n^2}right)x_n = 2 lim_{n to infty} x_n ] Since (x_n) grows exponentially, (y_n) will also grow exponentially, but the limit of the difference (x_{n+1} - x_n) will be finite. (blacksquare)