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

The Correct Option is D

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 5_n 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