Question

12．已知數(shù)列{a_n}為等差數(shù)列，且${a_{2015}}+{a_{2017}}=\int_0^2{\sqrt{4-{x^2}}}dx$，則a₂₀₁₆（a₂₀₁₄+a₂₀₁₈）的最小值為$\frac{{π}^{2}}{2}$．

Answer 1

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

解答 解：∵數(shù)列{a_n}為等差數(shù)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×$\frac{π}{2}×\frac{π}{2}$=$\frac{{π}^{2}}{2}$．
故答案為：$\frac{{π}^{2}}{2}$．

點評 本題考查等差數(shù)列的幾項和與積的最小值的求法，是中檔題，解題時要認真審題，注意定積分的性質(zhì)的合理運用．