# CAT 2021 Set-1 | Quantitative Aptitude | Question: 14

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If $x_{0} = 1, x_{1} = 2$ and $x_{n+2} = \dfrac{1 + x_{n+1}}{x_{n}}, n = 0, 1, 2, 3, \dots ,$ then $x_{2021}$ is equal to

1. $1$
2. $3$
3. $4$
4. $2$

Given that, $x_{0}=1, x_{1}=2, x_{n+2} = \frac{1+x_{n+1}}{x_{n}}; n=0,1,2,3,\dots$

Now,

• $x_{0}=1$
• $x_{1}=2$
• $x_{2}=\frac{1+2}{1} = 3$
• $x_{3}=\frac{1+3}{2} = 2$
• $x_{4}=\frac{1+2}{3} = 1$
• $x_{5}=\frac{1+1}{2} = 1$
• $x_{6}=\frac{1+1}{1} = 2$
• $x_{7}=\frac{1+2}{1} = 3$
• $x_{8}=\frac{1+3}{2} = 2$
• $x_{9}=\frac{1+2}{3} = 1$
• $\vdots \quad \vdots \quad \vdots \quad \vdots$

We can see the pattern,  $1,2,3,2,{\color{Red} {1}},1,2,3,2,{\color{Red} {1}},\dots$

Every $5^{\text{th}}$ multiple is $1.$

So, $x_{2020}=1.$

$\therefore$ The value of $x_{2021} = 2.$

Correct Answer $:\text{D}$

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