MathDB

4

Part of 2005 China National Olympiad

Problems(1)

fractional sequence inequality

Source: china mathematical olympiad cmo 2005 final round - Problem 4

1/25/2005
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}.
inequalitiescalculusfunctionderivativelogarithmsalgebra unsolvedalgebra