The sequence of positive numbers u₁, u₂, u₃, ... is such that u₁ < 5 and, for n ≥ 1, u(n+1) = (6u(n) + 5) / (u(n) + 2)

Mark scheme

Marking points from the official mark scheme (Oct/Nov 2025). The mark or mark type is shown in brackets.

1. [B1] States base case. (u₁ < 5 (given))
2. [B1] States inductive hypothesis. (Assume that u_k < 5 for some positive integer k.)
3. [M1] Considers 5-u_{k+1}, puts over common denominator. (Then 5-u_{k+1} = 5 - (6u_k + 5)/(u_k + 2) = (5u_k + 10 - 6u_k - 5)/(u_k + 2) = (5 - u_k)/(u_k + 2))
4. [A1] (5 - u_k)/(u_k + 2) > 0.
5. [A1] Hence, by induction, u_n < 5 for all positive integers n.
6. [M1 A1] Considers u_{n+1} - u_n or uses (5 - u_{n+1} = (5 - u_n)/(u_n + 2)). (u_{n+1} - u_n = (6u_n + 5)/(u_n + 2) - u_n = (6u_n + 5 - u_n² - 2u_n)/(u_n + 2) = (-u_n² + 4u_n + 5)/(u_n + 2))
7. [A1] Uses u_n < 5. ((-u_n² + 4u_n + 5)/(u_n + 2) = (-(u_n - 5)(u_n + 1))/(u_n + 2)). So u_n < 5 implies u_{n+1} - u_n > 0.