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Câu hỏi: Cho số thực a và dãy số ( ${{u}_{n}}$ ) được xác định bởi ${{u}_{1}}=a;{{u}_{n+1}}=1+\dfrac{{{u}_{n}}}{2}$. Khi đó $L=\lim {{u}_{n}}$ bằng
A. $L=1+2a.$
B. $L=2.$
C. $L=1.$
D. $L=3.$
A. $L=1+2a.$
B. $L=2.$
C. $L=1.$
D. $L=3.$
$\left\{ \begin{aligned}
& {{u}_{2}}=1+\dfrac{a}{2} \\
& {{u}_{3}}=1+\dfrac{{{u}_{2}}}{2}=1+\dfrac{1}{2}+\dfrac{a}{{{2}^{2}}} \\
& {{u}_{4}}=1+\dfrac{{{u}_{3}}}{2}=1+\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{a}{{{2}^{3}}} \\
& ........................ \\
\end{aligned} \right.\Rightarrow {{u}_{n}}=\left[ 1+\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+...+\dfrac{1}{{{2}^{n-2}}} \right]+\dfrac{a}{{{2}^{n-1}}} $ (với $ n\ge 3$)
$\Rightarrow {{u}_{n}}=1.\dfrac{1-{{\left( \dfrac{1}{2} \right)}^{n-1}}}{1-\dfrac{1}{2}}+\dfrac{a}{{{2}^{n-1}}}=2-\dfrac{1}{{{2}^{n-2}}}+\dfrac{a}{{{2}^{n-1}}}.$
$L=\lim {{u}_{n}}=\lim \left( 2-\dfrac{1}{{{2}^{n-2}}}+\dfrac{a}{{{2}^{n-1}}} \right)=2.$
& {{u}_{2}}=1+\dfrac{a}{2} \\
& {{u}_{3}}=1+\dfrac{{{u}_{2}}}{2}=1+\dfrac{1}{2}+\dfrac{a}{{{2}^{2}}} \\
& {{u}_{4}}=1+\dfrac{{{u}_{3}}}{2}=1+\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{a}{{{2}^{3}}} \\
& ........................ \\
\end{aligned} \right.\Rightarrow {{u}_{n}}=\left[ 1+\dfrac{1}{2}+\dfrac{1}{{{2}^{2}}}+\dfrac{1}{{{2}^{3}}}+...+\dfrac{1}{{{2}^{n-2}}} \right]+\dfrac{a}{{{2}^{n-1}}} $ (với $ n\ge 3$)
$\Rightarrow {{u}_{n}}=1.\dfrac{1-{{\left( \dfrac{1}{2} \right)}^{n-1}}}{1-\dfrac{1}{2}}+\dfrac{a}{{{2}^{n-1}}}=2-\dfrac{1}{{{2}^{n-2}}}+\dfrac{a}{{{2}^{n-1}}}.$
$L=\lim {{u}_{n}}=\lim \left( 2-\dfrac{1}{{{2}^{n-2}}}+\dfrac{a}{{{2}^{n-1}}} \right)=2.$
Đáp án B.