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Câu hỏi: Cho cấp số cộng $\left( {{u}_{n}} \right)$ có tất cả các số hạng đều dương và thỏa mãn
A. $\dfrac{1}{2}.$
B. 3.
C. 0.
D. 2.
${{u}_{1}}+{{u}_{2}}+...+{{u}_{2018}}=4\left( {{u}_{1}}+{{u}_{2}}+...+{{u}_{1009}} \right).$
Giá trị nhỏ nhất của biểu thức $P=\log _{7}^{2}{{u}_{4}}+\log _{7}^{2}{{u}_{25}}+\dfrac{5}{2}$ bằngA. $\dfrac{1}{2}.$
B. 3.
C. 0.
D. 2.
Ta có ${{u}_{1}}+{{u}_{2}}+...+{{u}_{2018}}=4\left( {{u}_{1}}+{{u}_{2}}+...+{{u}_{1009}} \right)$
$\Leftrightarrow \dfrac{2018\left( {{u}_{1}}+{{u}_{2018}} \right)}{2}=4.\dfrac{1009\left( {{u}_{1}}+{{u}_{1009}} \right)}{2}\Leftrightarrow 1009\left( 2{{u}_{1}}+2017d \right)=2018\left( 2{{u}_{1}}+1008d \right)$
$\Leftrightarrow 2018{{u}_{1}}=1009d\Leftrightarrow d=2{{u}_{1}}.$ Ta có ${{u}_{n}}={{u}_{1}}+\left( n-1 \right)d,\left( n\ge 2 \right)$
Suy ra $\left\{ \begin{aligned}
& {{u}_{4}}={{u}_{1}}=3.2{{u}_{1}}=7{{u}_{1}} \\
& {{u}_{25}}={{u}_{1}}+24.2{{u}_{1}}=49{{u}_{1}}. \\
\end{aligned} \right.$
Khi đó $P=\log _{7}^{2}{{u}_{4}}+\log _{7}^{2}{{u}_{25}}+\dfrac{5}{2}={{\left( {{\log }_{7}}7u \right)}^{2}}+{{\left( {{\log }_{7}}49{{u}_{1}} \right)}^{2}}+\dfrac{5}{2}$
= ${{\left( 1+{{\log }_{7}}{{u}_{1}} \right)}^{2}}+{{\left( 2+{{\log }_{7}}{{u}_{1}} \right)}^{2}}+\dfrac{5}{2}=2{{\left( {{\log }_{7}}{{u}_{1}} \right)}^{2}}+6{{\log }_{7}}{{u}_{1}}+5+\dfrac{5}{2}=2{{\left( {{\log }_{7}}{{u}_{1}}+\dfrac{3}{2} \right)}^{2}}+3\ge 3.$
Vậy ${{P}_{\min }}=3,$ đạt được khi ${{\log }_{7}}{{u}_{1}}=-\dfrac{3}{2}\Leftrightarrow {{u}_{1}}=\dfrac{1}{7\sqrt{7}}\Rightarrow d=\dfrac{2}{7\sqrt{7}}.$
$\Leftrightarrow \dfrac{2018\left( {{u}_{1}}+{{u}_{2018}} \right)}{2}=4.\dfrac{1009\left( {{u}_{1}}+{{u}_{1009}} \right)}{2}\Leftrightarrow 1009\left( 2{{u}_{1}}+2017d \right)=2018\left( 2{{u}_{1}}+1008d \right)$
$\Leftrightarrow 2018{{u}_{1}}=1009d\Leftrightarrow d=2{{u}_{1}}.$ Ta có ${{u}_{n}}={{u}_{1}}+\left( n-1 \right)d,\left( n\ge 2 \right)$
Suy ra $\left\{ \begin{aligned}
& {{u}_{4}}={{u}_{1}}=3.2{{u}_{1}}=7{{u}_{1}} \\
& {{u}_{25}}={{u}_{1}}+24.2{{u}_{1}}=49{{u}_{1}}. \\
\end{aligned} \right.$
Khi đó $P=\log _{7}^{2}{{u}_{4}}+\log _{7}^{2}{{u}_{25}}+\dfrac{5}{2}={{\left( {{\log }_{7}}7u \right)}^{2}}+{{\left( {{\log }_{7}}49{{u}_{1}} \right)}^{2}}+\dfrac{5}{2}$
= ${{\left( 1+{{\log }_{7}}{{u}_{1}} \right)}^{2}}+{{\left( 2+{{\log }_{7}}{{u}_{1}} \right)}^{2}}+\dfrac{5}{2}=2{{\left( {{\log }_{7}}{{u}_{1}} \right)}^{2}}+6{{\log }_{7}}{{u}_{1}}+5+\dfrac{5}{2}=2{{\left( {{\log }_{7}}{{u}_{1}}+\dfrac{3}{2} \right)}^{2}}+3\ge 3.$
Vậy ${{P}_{\min }}=3,$ đạt được khi ${{\log }_{7}}{{u}_{1}}=-\dfrac{3}{2}\Leftrightarrow {{u}_{1}}=\dfrac{1}{7\sqrt{7}}\Rightarrow d=\dfrac{2}{7\sqrt{7}}.$
Đáp án B.