Question:hard

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: Jan 15, 2026
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The Correct Option is D

Solution and Explanation

The given recursive sequence is:

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

Step 1: Calculate initial terms to identify 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: Recognize the repeating pattern

The sequence exhibits a cycle of 5 terms. The pattern is as follows: - Terms of the form \( x_{5n} \) are 1, - Terms of the form \( x_{5n+1} \) are 2, - Terms of the form \( x_{5n+2} \) are 3.

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

To find \( x_{2021} \), we examine its position within the 5-term cycle: \[ 2021 \div 5 = 404 \text{ with a remainder of } 1 \] This remainder indicates that \( x_{2021} \) is equivalent to the term at position 1 in the cycle, which is \( x_1 \). Since \( x_1 = 2 \), \( x_{2021} \) is 2.

Final Answer:

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

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