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Câu hỏi: Cho dãy số $u\left( n \right)$ thỏa mãn ${{\log }^{3}}u_{1}^{2}-3\log {{u}_{5}}={{\log }^{3}}\left( {{u}_{2}}+9 \right)-\log u_{1}^{6}$ và ${{u}_{n+1}}={{u}_{n}}+3 \left( {{u}_{1}}>0 \right)$ với mọi $n\ge 1$. Đặt ${{S}_{n}}={{u}_{1}}+{{u}_{1}}+...+{{u}_{n}}$. Tìm giá trị nhỏ nhất của n để ${{S}_{n}}>\dfrac{5n}{2}+{{2018}^{2}}$.
A. 1647
B. 1650
C. 1648
D. 1165
A. 1647
B. 1650
C. 1648
D. 1165
Ta có: ${{u}_{n+1}}={{u}_{n}}+3 \left( {{u}_{1}}>0 \right)\Rightarrow {{u}_{n}}$ là cấp số cộng với công sai $d=3$.
Mặt khác: ${{\log }^{3}}u_{1}^{2}-3\log {{u}_{5}}={{\log }^{3}}\left( {{u}_{2}}+9 \right)-\log u_{1}^{6}$
$\Leftrightarrow {{\log }^{3}}u_{1}^{2}-3\log \left( {{u}_{1}}+4d \right)={{\log }^{3}}\left( {{u}_{1}}+d+9 \right)-\log u_{1}^{6}$
$\Leftrightarrow 8{{\log }^{3}}{{u}_{1}}-3\log \left( {{u}_{1}}+12 \right)={{\log }^{3}}\left( {{u}_{1}}+12 \right)-6\log {{u}_{1}}$
$\Leftrightarrow 8{{\log }^{3}}{{u}_{1}}+6\log {{u}_{1}}={{\log }^{3}}\left( {{u}_{1}}+12 \right)+3\log \left( {{u}_{1}}+12 \right)$
Xét hàm số $f\left( t \right)={{t}^{3}}+3t \left( t\in \mathbb{R} \right)$ ta có: ${f}'\left( t \right)=3{{t}^{2}}+1>0 \left( \forall t\in \mathbb{R} \right)\Rightarrow f\left( t \right)$ đồng biến trên $\mathbb{R}$
Khi đó $f\left( 2\log {{u}_{1}} \right)=f\left( \log \left( {{u}_{1}}+12 \right) \right)\Leftrightarrow 2\log {{u}_{1}}=\log \left( {{u}_{1}}+12 \right)$
$\Leftrightarrow u_{1}^{2}={{u}_{1}}+12\xrightarrow{{{u}_{1}}>0}{{u}_{1}}=4\Rightarrow {{S}_{n}}=\dfrac{{{u}_{1}}+{{u}_{n}}}{2}.n=\dfrac{4+4+3\left( n-1 \right)}{2}.n$
Ta có: ${{S}_{n}}>\dfrac{5n}{2}+{{2018}^{2}}\Leftrightarrow \dfrac{3n+5}{2}n>\dfrac{5n}{2}+{{2018}^{2}}\Leftrightarrow \dfrac{3{{n}^{2}}}{2}>{{2018}^{2}}\Leftrightarrow n>1647,7$
Do đó ${{n}_{\min }}=1648$.
Mặt khác: ${{\log }^{3}}u_{1}^{2}-3\log {{u}_{5}}={{\log }^{3}}\left( {{u}_{2}}+9 \right)-\log u_{1}^{6}$
$\Leftrightarrow {{\log }^{3}}u_{1}^{2}-3\log \left( {{u}_{1}}+4d \right)={{\log }^{3}}\left( {{u}_{1}}+d+9 \right)-\log u_{1}^{6}$
$\Leftrightarrow 8{{\log }^{3}}{{u}_{1}}-3\log \left( {{u}_{1}}+12 \right)={{\log }^{3}}\left( {{u}_{1}}+12 \right)-6\log {{u}_{1}}$
$\Leftrightarrow 8{{\log }^{3}}{{u}_{1}}+6\log {{u}_{1}}={{\log }^{3}}\left( {{u}_{1}}+12 \right)+3\log \left( {{u}_{1}}+12 \right)$
Xét hàm số $f\left( t \right)={{t}^{3}}+3t \left( t\in \mathbb{R} \right)$ ta có: ${f}'\left( t \right)=3{{t}^{2}}+1>0 \left( \forall t\in \mathbb{R} \right)\Rightarrow f\left( t \right)$ đồng biến trên $\mathbb{R}$
Khi đó $f\left( 2\log {{u}_{1}} \right)=f\left( \log \left( {{u}_{1}}+12 \right) \right)\Leftrightarrow 2\log {{u}_{1}}=\log \left( {{u}_{1}}+12 \right)$
$\Leftrightarrow u_{1}^{2}={{u}_{1}}+12\xrightarrow{{{u}_{1}}>0}{{u}_{1}}=4\Rightarrow {{S}_{n}}=\dfrac{{{u}_{1}}+{{u}_{n}}}{2}.n=\dfrac{4+4+3\left( n-1 \right)}{2}.n$
Ta có: ${{S}_{n}}>\dfrac{5n}{2}+{{2018}^{2}}\Leftrightarrow \dfrac{3n+5}{2}n>\dfrac{5n}{2}+{{2018}^{2}}\Leftrightarrow \dfrac{3{{n}^{2}}}{2}>{{2018}^{2}}\Leftrightarrow n>1647,7$
Do đó ${{n}_{\min }}=1648$.
Đáp án C.