A-Level Sequences and series: worked solution

Question

A sequence has u1=2 and u_{n+1}=1/(1−u_n) for n≥1. Find u2, u3 and u4, and hence give u2025.

The recurrence gives each term from the one before it, so generate the terms one at a time from u1 = 2. u2 = 1/(1 − u1) = 1/(1 − 2) = 1/(−1) = −1 u3 = 1/(1 − u2) = 1/(1 − (−1)) = 1/2 u4 = 1/(1 − u3) = 1/(1 − 1/2) = 1/(1/2) = 2 Now spot the pattern and justify it. u4 = 2 = u1, and every term depends only on the term immediately before it, so once a value repeats the whole sequence repeats. The sequence is periodic with period 3: 2, −1, 1/2, 2, −1, 1/2, ... So u_n depends only on the remainder when n is divided by 3: n ≡ 1 (mod 3) gives u_n = 2 n ≡ 2 (mod 3) gives u_n = −1 n ≡ 0 (mod 3) gives u_n = 1/2 Apply this to n = 2025. 2025 = 3 × 675, so 2025 is divisible by 3, i.e. 2025 ≡ 0 (mod 3). Therefore u2025 = u3 = 1/2.