Question

已知數列{a_n}中，a₁=1，a_n+3≤a_n+3，a_n+2≥a_n+2．求a₂₀₀₇．

Answer 1

分析：由a_n+2≥a_n+2，利用遞推關系可得：a₂₀₀₇≥a₂₀₀₅+2≥a₂₀₀₃+2×2≥…≥a₁+2×1003=2007．另一方面由   a_n+2≥a_n+2，得a_n≤a_n+2-2，可得a_n+3≤a_n+3≤a_n+2-2+3=a_n+2+1（n≥1）．于是 遞推下去可得：a₂₀₀₇≤a₂₀₀₆+1≤a₂₀₀₅+1×2≤a₂₀₀₂+3+1×2≤a₁₉₉₉+3×2+1×2≤…≤a₁+3×668+1×2=2007，利用“夾逼法”
即可得出a₂₀₀₇．

解答：解：由題設a_n+2≥a_n+2，可得a₂₀₀₇≥a₂₀₀₅+2≥a₂₀₀₃+2×2≥…≥a₁+2×1003=2007．
由 a_n+2≥a_n+2，得a_n≤a_n+2-2，則a_n+3≤a_n+3≤a_n+2-2+3=a_n+2+1（n≥1）．
于是   a₂₀₀₇≤a₂₀₀₆+1≤a₂₀₀₅+1×2≤a₂₀₀₂+3+1×2≤a₁₉₉₉+3×2+1×2≤…≤a₁+3×668+1×2=2007，
∴a₂₀₀₇=2007．

點評：本題考查了利用“夾逼法”和遞推式的意義即不等式的性質求數列的某一項，屬于難題．