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Câu hỏi: Dãy số $\left( {{u}_{n}} \right)$ được cho bởi $\left\{ \begin{aligned}
& {{u}_{1}}=2;{{u}_{2}}=5 \\
& {{u}_{n+2}}=6{{u}_{n+1}}-5{{u}_{n}} \\
\end{aligned} \right.\left( n\ge 1,n\in \mathbb{N} \right) $. Số nguyên n lớn nhất sao cho $ {{\log }_{2}}\left( {{u}_{n}} \right)<102$ là
A. 51.
B. 48.
C. 56.
D. 45.
& {{u}_{1}}=2;{{u}_{2}}=5 \\
& {{u}_{n+2}}=6{{u}_{n+1}}-5{{u}_{n}} \\
\end{aligned} \right.\left( n\ge 1,n\in \mathbb{N} \right) $. Số nguyên n lớn nhất sao cho $ {{\log }_{2}}\left( {{u}_{n}} \right)<102$ là
A. 51.
B. 48.
C. 56.
D. 45.
Ta có ${{u}_{n+2}}=6{{u}_{n+1}}-5{{u}_{n}}\Rightarrow {{u}_{n+2}}-{{u}_{n+1}}=5\left( {{u}_{n+1}}-{{u}_{n}} \right)$
Đặt ${{v}_{n}}={{u}_{n}}-{{u}_{n-1}}$ thì ${{v}_{n+1}}=5{{v}_{n}},$ với $n\ge 2.$
Suy ra ${{v}_{n}}=5{{v}_{n-1}}=5.\left( 5{{v}_{n-2}} \right)={{5}^{2}}.\left( 5{{v}_{n-3}} \right)=.......={{5}^{n-3}}.\left( 5{{v}_{2}} \right)={{5}^{n-2}}.\left( {{u}_{2}}-{{u}_{1}} \right)={{3.5}^{n-2}}$
Ta có ${{u}_{n}}=\left( {{u}_{n}}-{{u}_{n-1}} \right)+\left( {{u}_{n-1}}-{{u}_{n-2}} \right)+\left( {{u}_{n-2}}-{{u}_{n-3}} \right)+.....+\left( {{u}_{2}}-{{u}_{1}} \right)+{{u}_{1}}$
$={{v}_{n}}+{{v}_{n-1}}+{{v}_{n-2}}+....+{{v}_{2}}+{{u}_{1}}={{3.5}^{n-2}}+{{3.5}^{n-3}}+{{3.5}^{n-4}}+....+{{3.5}^{0}}+2$
$=2+3\left( 1+5+{{5}^{2}}+....+{{5}^{n-2}} \right)=2+3.\dfrac{1-{{5}^{n-1}}}{1-5}=\dfrac{{{3.5}^{n-1}}+5}{4}$
Do đó ${{\log }_{2}}\left( {{u}_{n}} \right)<102\Leftrightarrow {{u}_{n}}<{{2}^{102}}\Leftrightarrow \dfrac{{{3.5}^{n-1}}+5}{4}<{{2}^{102}}\Leftrightarrow {{5}^{n-1}}<\dfrac{{{2}^{104}}-5}{3}$
$\Rightarrow n-1<{{\log }_{5}}\dfrac{{{2}^{104}}-5}{3}\approx 44,1077.$
Vậy n = 45 là số cần tìm.
Đặt ${{v}_{n}}={{u}_{n}}-{{u}_{n-1}}$ thì ${{v}_{n+1}}=5{{v}_{n}},$ với $n\ge 2.$
Suy ra ${{v}_{n}}=5{{v}_{n-1}}=5.\left( 5{{v}_{n-2}} \right)={{5}^{2}}.\left( 5{{v}_{n-3}} \right)=.......={{5}^{n-3}}.\left( 5{{v}_{2}} \right)={{5}^{n-2}}.\left( {{u}_{2}}-{{u}_{1}} \right)={{3.5}^{n-2}}$
Ta có ${{u}_{n}}=\left( {{u}_{n}}-{{u}_{n-1}} \right)+\left( {{u}_{n-1}}-{{u}_{n-2}} \right)+\left( {{u}_{n-2}}-{{u}_{n-3}} \right)+.....+\left( {{u}_{2}}-{{u}_{1}} \right)+{{u}_{1}}$
$={{v}_{n}}+{{v}_{n-1}}+{{v}_{n-2}}+....+{{v}_{2}}+{{u}_{1}}={{3.5}^{n-2}}+{{3.5}^{n-3}}+{{3.5}^{n-4}}+....+{{3.5}^{0}}+2$
$=2+3\left( 1+5+{{5}^{2}}+....+{{5}^{n-2}} \right)=2+3.\dfrac{1-{{5}^{n-1}}}{1-5}=\dfrac{{{3.5}^{n-1}}+5}{4}$
Do đó ${{\log }_{2}}\left( {{u}_{n}} \right)<102\Leftrightarrow {{u}_{n}}<{{2}^{102}}\Leftrightarrow \dfrac{{{3.5}^{n-1}}+5}{4}<{{2}^{102}}\Leftrightarrow {{5}^{n-1}}<\dfrac{{{2}^{104}}-5}{3}$
$\Rightarrow n-1<{{\log }_{5}}\dfrac{{{2}^{104}}-5}{3}\approx 44,1077.$
Vậy n = 45 là số cần tìm.
Đáp án D.