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2023 Algebra NT #6 a_k =\frac{ka_{k-1}}{a_{k-1} - (k - 1)}

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February 28, 2024
algebra

Problem Statement

Suppose a1,a2,...,a100a_1, a_2, ... , a_{100} are positive real numbers such that ak=kak1ak1(k1)a_k =\frac{ka_{k-1}}{a_{k-1} - (k - 1)} for k=2,3,...,100k = 2, 3, ... , 100. Given that a20=a23a_{20} = a_{23}, compute a100a_{100}.