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TR-MO-R1 2017 Problem 22

    f(0)=23f(0)=\dfrac{2}{3} ve n=1,2,3,…n=1,2,3,\ldots için f(n)≠0f(n)\neq 0 ve (f(n+1)−1)(f(n)+3)+3=0\big(f(n+1)-1\big)\big(f(n)+3\big)+3=0 olduğuna göre

    1f(0)+1f(1)+1f(2)+1f(3)+⋯+1f(2016)+1f(2017)=?\frac{1}{f(0)}+\frac{1}{f(1)}+\frac{1}{f(2)}+\frac{1}{f(3)}+\cdots+\frac{1}{f(2016)}+\frac{1}{f(2017)}=?

    A) 32018−10103^{2018}-1010 B) 32017−10093^{2017}-1009 C) 2⋅32018−10092\cdot 3^{2018}-1009 D) 2⋅(32017−505)2\cdot\left(3^{2017}-505\right) E) 2⋅32017−10092\cdot 3^{2017}-1009

    Official text Turkish Mathematical Olympiad, first round 2017 official paper, page 13. Quoted with source.

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