Question:

If \(x_0 = 1, x_1 = 2, and \space x_{n + 2} =\frac{ 1+x_{n+1}}{x_n}, n = 0, 1, 2, 3,...,\) then \(x_{2021}\) is equal to?

Updated On: Jul 23, 2025
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The Correct Option is D

Solution and Explanation

We are given the following recursive sequence:

\[ x_0 = 1, \quad x_1 = 2, \quad x_{n+2} = \frac{1 + x_{n+1}}{x_n} \]

Step 1: Finding the first few terms to detect a pattern

- For \( n = 0 \): \[ x_2 = \frac{1 + x_1}{x_0} = \frac{1 + 2}{1} = 3 \] - For \( n = 1 \): \[ x_3 = \frac{1 + x_2}{x_1} = \frac{1 + 3}{2} = 2 \] - For \( n = 2 \): \[ x_4 = \frac{1 + x_3}{x_2} = \frac{1 + 2}{3} = 1 \] - For \( n = 3 \): \[ x_5 = \frac{1 + x_4}{x_3} = \frac{1 + 1}{2} = 1 \] - For \( n = 4 \): \[ x_6 = \frac{1 + x_5}{x_4} = \frac{1 + 1}{1} = 2 \]

Step 2: Identifying the repeating pattern

The sequence enters a repetitive pattern every 5 terms. Specifically: - Terms at positions \( 5n \) are 1, - Terms at positions \( 5n+1 \) are 2, - Terms at positions \( 5n+2 \) are 3, and so on.

Step 3: Finding \( x_{2021} \)

We need to determine the term corresponding to position 2021. Since the pattern repeats every 5 terms: \[ 2021 \div 5 = 404 \text{ remainder } 1 \] Therefore, \( x_{2021} \) corresponds to \( x_1 \), which is 2.

Final Answer:

The correct value of \( x_{2021} \) is \( \boxed{2} \).

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