The sequence xn is given by the formula x_n = n^3 − 9n^2 + 631. What is the largest value of n for which x_n > x_(n+1)?

We know that x_n > x_(n+1) is true if and only if x_n - x_(n+1) > 0 is true.So x_n - x_(n+1) = (n^3 − 9n^2 + 631) − ((n + 1)^3 − 9(n + 1)^2 + 631) = (n^3 − n^3 − 3n^2 − 3n − 1) − 9(n^2 − n^2 − 2n − 1) = −3n^2 + 15n + 8We know how to solve quadratic equations hence we see that −3x^2 + 15x + 8 = 0 if x = (15 - sqrt(321))/6 or x = (15 + sqrt(321))/6So −3x^2 + 15x + 8 > 0 precisely when (15 - sqrt(321))/6 < x < (15 + sqrt(321))/6. (To see this think of the graph of f(x) = −3x^2 + 15x + 8, which is a parabola and think what part of this parabola is above the x-axis so that f(x) is positive)Now noting that 18^2 = 324 we see that sqrt(321) is approximately 18 and so (15 + sqrt(321))/6 is approximately 5.5. Obviously, the largest integer which is smaller that 5.5 is 5.Hence, the answer is 5.