If $\mathrm{x}_{0}=1, \mathrm{x}_{1}=2$, and $\mathrm{x}_{\mathrm{n}+2}=\frac{1+x_{n+1}}{x_{n}}, n=0,1,2,3 \ldots \ldots$., then $x_{2021}$ is

Given:

$x_0 = 1$

$x_1 = 2$

$x_{n+2} = \tfrac{1+x_{n+1}}{x_n}$

Let us calculate the next few terms:

$x_2 = \tfrac{1+x_1}{x_0} = \tfrac{1+2}{1} = 3$

$x_3 = \tfrac{1+x_2}{x_1} = \tfrac{1+3}{2} = 2$

$x_4 = \tfrac{1+x_3}{x_2} = \tfrac{1+2}{3} = 1$

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Let us continue calculating to see if there's a repeating pattern:

$x_5 = \tfrac{1+x_4}{x_3} = \tfrac{1+1}{2} = 1$

$x_6 = \tfrac{1+x_5}{x_4} = \tfrac{1+1}{1} = 2$

$x_7 = \tfrac{1+x_6}{x_5} = \tfrac{1+2}{1} = 3$

$x_8 = \tfrac{1+x_7}{x_6} = \tfrac{1+3}{2} = 2$

$x_9 = \tfrac{1+x_8}{x_7} = \tfrac{1+2}{3} = 1$

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The sequence is: $1, 2, 3, 2, 1, 1, 2, 3, 2, 1, 1, 2, 3, 2, 1, ...$

Starting from $x_0$, the sequence repeats every 5 terms:

$x_0 = 1$

$x_1 = 2$

$x_2 = 3$

$x_3 = 2$

$x_4 = 1$

Then it repeats:

$x_5 = 1$ (same as $x_0$)

$x_6 = 2$ (same as $x_1$)

$x_7 = 3$ (same as $x_2$)

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Since the sequence has a period of 5, we need to find the remainder when 2021 is divided by 5:

$2021 = 5 \times 404 + 1$

The remainder is 1, which means $x_{2021} = x_1 = 2$.

(Please note that in a periodic sequence, the index determines the term, not the position within a completed block. If the period is $5$, then $x_n = x_{n \bmod 5}.$ Thus, $2020 \equiv 0 \pmod{5}\Rightarrow x_{2020}=x_0$, while $2021 \equiv 1 \pmod{5}\Rightarrow x_{2021}=x_1$.

$= x_1 = 2$.