[tex]\it S\breve{a}\ se\ determine \ termenii \ x_2, \ x_3,\ x_4 \ \c{s}i\ s\breve{a}\ se \ exprime\ x_{n+2}
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{\^{i}n\ func\c{\it t}ie\ de\ x_n,\ dac\breve{a}: \ x_1=4,\ x_n=x_{n-1} \cdot \dfrac{n}{n + 2}.[/tex]
Rezolvare:
Din enunț se cunosc două relații :
[tex]\it x_1=4 \ \ \ \ (*)
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\ x_n=x_{n-1} \cdot \dfrac{n}{n - 2} \ \ \ \ \ (**)[/tex]
[tex]\it n=2 \stackrel{(**)}{\Longrightarrow} x_2=x_{2-1} \cdot\dfrac{2}{2+2}=x_1\cdot\dfrac{2}{4} \stackrel{(*)}{=} 4\cdot\dfrac{2}{4} =2 \\\;\\ \\\;\\ n=3 \stackrel{(**)}{\Longrightarrow} x_3=x_{3-1} \cdot\dfrac{3}{3+2}=x_2\cdot\dfrac{3}{5} =2\cdot\dfrac{3}{5} = \dfrac{6}{5}
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[tex]\it n=4 \stackrel{(**)}{\Longrightarrow} x_4=x_{4-1} \cdot\dfrac{4}{4+2}=x_3\cdot\dfrac{4}{6} =\dfrac{6}{5}\cdot\dfrac{4}{6} = \dfrac{4}{5}
[/tex]
[tex]\it (**) \Rightarrow x_{n+1} =x_{n+1-1}\cdot\dfrac{n+1}{n+1+2} = x_n\cdot\dfrac{n+1}{n+3}
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(**) \Rightarrow x_{n+2} =x_{n+2-1}\cdot\dfrac{n+2}{n+2+2} = x_{n+1}\cdot\dfrac{n+2}{n+4} [/tex]
