The sequence \(x_n\) is given by the formula \[x_n=n^3-9n^2+631.\]
The largest value of \(n\) for which \(x_n>x_{n+1}\) is
5;
7;
11;
17.
We see that \(x_n>x_{n+1}\) is equivalent to saying \(x_n-x_{n+1}>0\).
So we solve this inequality for \(n\): we have \[\begin{align*} & n^3-9n^2+631-[(n+1)^3-9(n+1)^2+631] > 0 \\ & \iff n^3-9n^2+631-(n^3+3n^2+3n+1-9n^2-18n-9+631) > 0 \\ & \iff -3n^2+15n+8 > 0 \\ & \iff 3n^2-15n-8 < 0. \end{align*}\] Using the quadratic formula, the solution to \(3n^2-15n-8=0\) is \[\begin{align*} n &= \frac{15\pm\sqrt{15^2-4\times 3\times(-8)}}{6} \\ &= \frac{15\pm\sqrt{225+96}}{6} \\ &= \frac{15\pm\sqrt{321}}{6}. \end{align*}\]So the values of \(n\) satisfying the inequality are \[\frac{15-\sqrt{321}}{6} < n < \frac{15+\sqrt{321}}{6} \approx 5.49.\] This implies that \(n=5\) is the largest integer in the range.
Hence the answer required is (a).
Check: \(x_4 =551, x_5 =531 , x_6 = 523, x_7= 533.\)
