Maths Olympiad Prep

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Algebra Difficulty 5.3 AIME, harder Prove it Philippines

Problem:
Let {an}\{a_n\} be a sequence such that a1=20a_1 = 20, a2=17a_2 = 17 and an+2an−2=3an−1a_n + 2 a_{n-2} = 3 a_{n-1}. Determine the value of a2017−a2016a_{2017} - a_{2016}.

Solution

Solution:
We have an=3an−1−2an−2⟹an−an−1=2(an−1−an−2)a_n = 3 a_{n-1} - 2 a_{n-2} \Longrightarrow a_n - a_{n-1} = 2(a_{n-1} - a_{n-2}). Repeated use of this recurrence relation gives a2017−a2016=22015(a2−a1)=−3⋅22015a_{2017} - a_{2016} = 2^{2015}(a_2 - a_1) = -3 \cdot 2^{2015}.

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