Review question

# What is the largest value of $n$ for which $x_n>x_{n+1}$? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9473

## Solution

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

1. 5;

2. 7;

3. 11;

4. 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: $x_4 =551, x_5 =531 , x_6 = 523, x_7= 533.$