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ASU 015 All Russian MO d 1962 8.3 a_{100}=3a_{99}-2a_{98}

Source:

June 17, 2019
algebra

Problem Statement

Given positive numbers a1,a2,...,a99,a100a_1,a_2,...,a_{99},a_{100}. It is known, that a1>a0,a2=3a12a0,a3=3a22a1,...,a100=3a992a98a_1>a_0, a_2=3a_1-2a_0, a_3=3a_2-2a_1, ..., a_{100}=3a_{99}-2a_{98} Prove that a100>299.a_{100}>2^{99}.