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?

• A. 4
• B. 3
• C. 1
• D. 2

Answer 1

Answer: (D)

2

Answer 2

Given that:
$x_0 = 1$
$x_1 = 2$
$x_{n+2} = \frac{1+x_{n+1}}{x_n}$

Let's find the next few terms to determine a pattern:

Put $n=0$
⇒ $x_2 = \frac{1 + x_1}{x_0} = \frac{1 + 2}{1} = 3$

Put $n=1$
⇒ $x_3 = \frac{1 + x_2}{x_1} = \frac{1 + 3}{2} = 2$

Put $n=2$
⇒ $x_4 = \frac{1 + x_3}{x_2} = \frac{1 + 2}{3} = 1$

Put $n=3$
⇒ $x_5 = \frac{1 + x_4}{x_3} = \frac{1 + 1}{2} = 1$

Put $n=4$
⇒ $x_6 = \frac{1 + x_5}{x_4} = \frac{1 + 1}{1} = 2$

Therefore, the series enters a repetitive pattern every 5 terms. The terms corresponding to numbers in the format 5n are 1, those in the format $5_{n+1}$ are 2, and so forth, with n taking values from 0, 1, 2, 3, and so on.
Since 2021 can be expressed as $5_{n+1},$ its value will be 2.

So, the correct option is (D): 2