Câu hỏi của Nguyễn Thu Phương

Question

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Nguyễn Việt Lâm · Accepted Answer

$u_{n+1}=u_n+\dfrac{1}{n\left(n+1\right)}\Rightarrow u_{n+1}=u_n+\dfrac{1}{n}-\dfrac{1}{n+1}$

$\Rightarrow u_{n+1}+\dfrac{1}{n+1}=u_n+\dfrac{1}{n}$

Đặt $u_n+\dfrac{1}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1+\dfrac{1}{1}=2\v_{n+1}=v_n\end{matrix}\right.$

Từ $v_{n+1}=v_n\Rightarrow v_{n+1}=v_n=v_{n-1}=...=v_1=2$

$\Rightarrow v_n=2\Rightarrow u_n+\dfrac{1}{n}=2$

$\Rightarrow u_n=2-\dfrac{1}{n}=\dfrac{2n-1}{n}$

$\Rightarrow u_{2024}=\dfrac{2.2024-1}{2024}=\dfrac{4047}{2024}$

Câu trả lời:

$u_{n+1}=u_n+\dfrac{1}{n\left(n+1\right)}\Rightarrow u_{n+1}=u_n+\dfrac{1}{n}-\dfrac{1}{n+1}$

$\Rightarrow u_{n+1}+\dfrac{1}{n+1}=u_n+\dfrac{1}{n}$

Đặt $u_n+\dfrac{1}{n}=v_n\Rightarrow\left\{{}\begin{matrix}v_1=u_1+\dfrac{1}{1}=2\v_{n+1}=v_n\end{matrix}\right.$

Từ $v_{n+1}=v_n\Rightarrow v_{n+1}=v_n=v_{n-1}=...=v_1=2$

$\Rightarrow v_n=2\Rightarrow u_n+\dfrac{1}{n}=2$

$\Rightarrow u_n=2-\dfrac{1}{n}=\dfrac{2n-1}{n}$

$\Rightarrow u_{2024}=\dfrac{2.2024-1}{2024}=\dfrac{4047}{2024}$

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