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0
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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
0

$a_{100} = \boxed{1/6}$

Jun 13, 2021
#2
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Can you send working?

Guest Jun 13, 2021
#3
+26319
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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 $$a_{100}$$

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

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

Jun 13, 2021