Bài tập nâng cao · Bài 3
Bài tập nâng cao Dãy số · Bài 3
Cho dãy số $(u_n)$ biết $\left\{ \begin{array}{l} u_1=1 \\ u_{n+1}=\frac{u_n+2}{u_n+1} \end{array} \right.$. Tìm số hạng $u_{10}$.
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Lời giải
$u_2=\frac{u_1+2}{u_1+1}=\frac{1+2}{1+1}=\frac{3}{2}$ ; $u_3=\frac{u_2+2}{u_2+1}=\frac{\frac{3}{2}+2}{\frac{3}{2}+1}=\frac{7}{5}$ ; $u_4=\frac{u_3+2}{u_3+1}=\frac{\frac{7}{5}+2}{\frac{7}{5}+1}=\frac{17}{12}$ ; $u_5=\frac{u_4+2}{u_4+1}=\frac{\frac{17}{12}+2}{\frac{17}{12}+1}=\frac{41}{29}$ ; $u_6=\frac{u_5+2}{u_5+1}=\frac{\frac{41}{29}+2}{\frac{41}{29}+1}=\frac{99}{70}$ ; $u_7=\frac{u_6+2}{u_6+1}=\frac{\frac{99}{70}+2}{\frac{99}{70}+1}=\frac{239}{169}$
$u_8=\frac{u_7+2}{u_7+1}=\frac{\frac{239}{169}+2}{\frac{239}{169}+1}=\frac{577}{408}$ ; $u_9=\frac{u_8+2}{u_8+1}=\frac{\frac{577}{408}+2}{\frac{577}{408}+1}=\frac{1393}{985}$ ; $u_{10}=\frac{u_9+2}{u_9+1}=\frac{\frac{1393}{985}+2}{\frac{1393}{985}+1}=\frac{3363}{2378}$
$u_8=\frac{u_7+2}{u_7+1}=\frac{\frac{239}{169}+2}{\frac{239}{169}+1}=\frac{577}{408}$ ; $u_9=\frac{u_8+2}{u_8+1}=\frac{\frac{577}{408}+2}{\frac{577}{408}+1}=\frac{1393}{985}$ ; $u_{10}=\frac{u_9+2}{u_9+1}=\frac{\frac{1393}{985}+2}{\frac{1393}{985}+1}=\frac{3363}{2378}$