Question

12．已知数列{a_n}为等差数列，且${a_{2015}}+{a_{2017}}=\int_0^2{\sqrt{4-{x^2}}}dx$，则a₂₀₁₆（a₂₀₁₄+a₂₀₁₈）的最小值为$\frac{{π}^{2}}{2}$．

Answer 1

分析 先求出2a₂₀₁₆=${a_{2015}}+{a_{2017}}=\int_0^2{\sqrt{4-{x^2}}}dx$=π，进而a₂₀₁₆=$\frac{π}{2}$，由此能求出a₂₀₁₆（a₂₀₁₄+a₂₀₁₈）的值．

解答 解：∵数列{a_n}为等差数列，且${a_{2015}}+{a_{2017}}=\int_0^2{\sqrt{4-{x^2}}}dx$，
∴2a₂₀₁₆=${a_{2015}}+{a_{2017}}=\int_0^2{\sqrt{4-{x^2}}}dx$=$\frac{1}{4}$×π×2²=π，
∴a₂₀₁₆=$\frac{π}{2}$，
a₂₀₁₆（a₂₀₁₄+a₂₀₁₈）=2a₂₀₁₆•a₂₀₁₆=2×$\frac{π}{2}×\frac{π}{2}$=$\frac{{π}^{2}}{2}$．
故答案为：$\frac{{π}^{2}}{2}$．

点评 本题考查等差数列的几项和与积的最小值的求法，是中档题，解题时要认真审题，注意定积分的性质的合理运用．