olymmath

CMO 2005 Problem 4

  • Day 2

The sequence {an}\{a_n\} is defined by: a1=2116a_1=\frac{21}{16} , and for n≥2n\ge2 ,2an−3an−1=32n+1.2a_n-3a_{n-1}=\frac{3}{2^{n+1}}.Let mm be an integer with m≥2m\ge2 . Prove that: for n≤mn\le m , we have(an+32n+3)1m(m−(23)n(m−1)m)<m2−1m−n+1.\left(a_n+\frac{3}{2^{n+3}}\right)^{\frac{1}{m}}\left(m-\left(\frac{2}{3}\right)^{{\frac{n(m-1)}{m}}}\right)<\frac{m^2-1}{m-n+1}.

Dataset text AI-MO AoPS. Quoted with source.

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