What is the largest value of n n n x n > x n + 1 x_n>x_{n+1} x n > x n + 1
Question
The sequence x n x_n x n x n = n 3 − 9 n 2 + 631. x_n=n^3-9n^2+631. x n = n 3 − 9 n 2 + 631.
The largest value of n n n x n > x n + 1 x_n>x_{n+1} x n > x n + 1
5;
7;
11;
Suggestion
How else could we write x n > x n + 1 x_n>x_{n+1} x n > x n + 1 n n n
Solution
We see that x n > x n + 1 x_n>x_{n+1} x n > x n + 1 x n − x n + 1 > 0 x_n-x_{n+1}>0 x n − x n + 1 > 0
So we solve this inequality for
n n n
: we have
n 3 − 9 n 2 + 631 − [ ( n + 1 ) 3 − 9 ( n + 1 ) 2 + 631 ] > 0 ⟺ n 3 − 9 n 2 + 631 − ( n 3 + 3 n 2 + 3 n + 1 − 9 n 2 − 18 n − 9 + 631 ) > 0 ⟺ − 3 n 2 + 15 n + 8 > 0 ⟺ 3 n 2 − 15 n − 8 < 0. \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*} n 3 − 9 n 2 + 631 − [( n + 1 ) 3 − 9 ( n + 1 ) 2 + 631 ] > 0 ⟺ n 3 − 9 n 2 + 631 − ( n 3 + 3 n 2 + 3 n + 1 − 9 n 2 − 18 n − 9 + 631 ) > 0 ⟺ − 3 n 2 + 15 n + 8 > 0 ⟺ 3 n 2 − 15 n − 8 < 0. Using the quadratic formula, the solution to
3 n 2 − 15 n − 8 = 0 3n^2-15n-8=0 3 n 2 − 15 n − 8 = 0
is
n = 15 ± 15 2 − 4 × 3 × ( − 8 ) 6 = 15 ± 225 + 96 6 = 15 ± 321 6 . \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*} n = 6 15 ± 1 5 2 − 4 × 3 × ( − 8 ) = 6 15 ± 225 + 96 = 6 15 ± 321 . So the values of n n n 15 − 321 6 < n < 15 + 321 6 ≈ 5.49. \frac{15-\sqrt{321}}{6} < n < \frac{15+\sqrt{321}}{6} \approx 5.49. 6 15 − 321 < n < 6 15 + 321 ≈ 5.49. n = 5 n=5 n = 5
Hence the answer required is (a).
Check: x 4 = 551 , x 5 = 531 , x 6 = 523 , x 7 = 533. x_4 =551, x_5 =531 , x_6 = 523, x_7= 533. x 4 = 551 , x 5 = 531 , x 6 = 523 , x 7 = 533.