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The sequence (an) satisfies a1=3 and

 

\({a}_{n}=\frac{{a}_{n-1}-1}{{a}_{n-1}+1}\)

 

for all n >= 2. Find a100

 Jun 13, 2021
 #1
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$a_{100} = \boxed{1/6}$

 Jun 13, 2021
 #2
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Can you send working?

Guest Jun 13, 2021
 #3
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The sequence \((a_n)\) satisfies \(a_1=3\) and
\(a_{n}=\dfrac{a_{n-1}-1}{a_{n-1}+1}\)
for all \(n \ge 2\). Find \(a_{100}\)

 

\(\begin{array}{|rclclcr|} \hline a_1 && && &=& \color{red}3 \\\\ a_2 &=& \dfrac{a_1-1}{a_1+1} &=& \dfrac{3-1}{3+1}&=& \dfrac{1}{2} \\\\ a_3 &=& \dfrac{a_2-1}{a_2+1} &=& \dfrac{\dfrac{1}{2}-1}{\dfrac{1}{2}+1}&=& -\dfrac{1}{3} \\\\ a_4 &=& \dfrac{a_3-1}{a_3+1} &=& \dfrac{-\dfrac{1}{3}-1}{-\dfrac{1}{3}+1}&=& -2 \\\\ a_5 &=& \dfrac{a_4-1}{a_4+1} &=& \dfrac{-2-1}{-2+1}&=& \color{red}3 \\\\ a_6 &&&=& a_2 &=& \dfrac{1}{2} \\\\ a_7 &&&=& a_3 &=& -\dfrac{1}{3} \\\\ a_8 &&&=& a_4 &=& -2\\\\ a_9 &&&=& a_5 &=& \color{red}3 \\ \dots \\ \hline \end{array}\)

 

\(\mathbf{100} \text{ is divisible by } 4 \text{, so } a_{100} = a_{8} = a_{4} = \mathbf{-2}\)

 

laugh

 Jun 13, 2021

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