We are given the following recursive sequence:
\[ x_0 = 1, \quad x_1 = 2, \quad x_{n+2} = \frac{1 + x_{n+1}}{x_n} \]
- 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 \( n = 4 \): \[ x_6 = \frac{1 + x_5}{x_4} = \frac{1 + 1}{1} = 2 \]
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.
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.
The correct value of \( x_{2021} \) is \( \boxed{2} \).
LIST I | LIST II | ||
A. | The solution set of the inequality \(-5x > 3, x\in R\), is | I. | \([\frac{20}{7},∞)\) |
B. | The solution set of the inequality is, \(\frac{-7x}{4} ≤ -5, x\in R\) is, | II. | \([\frac{4}{7},∞)\) |
C. | The solution set of the inequality \(7x-4≥0, x\in R\) is, | III. | \((-∞,\frac{7}{5})\) |
D. | The solution set of the inequality \(9x-4 < 4x+3, x\in R\) is, | IV. | \((-∞,-\frac{3}{5})\) |