ChallengeSuperchallengeAdvanced Number Theory ShareAdvanced Number Theory · Superchallenge problemA sequence is defined by a1=3a_1=3a1=3 and an+1=11−ana_{n+1}=\dfrac{1}{1-a_n}an+1=1−an1 for n≥1n\ge1n≥1. Then a2025=mna_{2025}=\dfrac{m}{n}a2025=nm in lowest terms. Find m+nm+nm+n.Check Show answer Walk through it step by step Practice more like this Review this concept