Question

已知数列{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，利用递推关系可得：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 递推下去可得：a₂₀₀₇≤a₂₀₀₆+1≤a₂₀₀₅+1×2≤a₂₀₀₂+3+1×2≤a₁₉₉₉+3×2+1×2≤…≤a₁+3×668+1×2=2007，利用“夹逼法”
即可得出a₂₀₀₇．

解答：解：由题设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，则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．

点评：本题考查了利用“夹逼法”和递推式的意义即不等式的性质求数列的某一项，属于难题．