$\left\{\begin{matrix} u_{1}=\frac{1}{2} & \\ \frac{u_{n}-u_{n+1}}{u_{n+1}}=u_{n}+\sqrt{u_{n}^{2}+4u_{n}}& \end{matrix}\right.$
đặt $S_{n}=\sum_{k=1}^{n}u_{n}$. Tìm phần nguyên [$S_{n}$]
#2
Posted 26-08-2018 - 03:32
Hr MiSu
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bằng quy nạp ta có: $u_n=\frac{1}{n(n+1)}\Rightarrow \left \lfloor S_n \right \rfloor=1$
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