CAT 2021 Slot 1 QA Question 11: If $x_0 = 1, x_1 = 2,\ ext{and}\ x_{n+2} = \dfrac{1 + x_{n+1}}{x_n},\ n = 0,1,2,3,\ldots$, then $x_{2021}$ is equal to

Question

If $x_0 = 1, x_1 = 2,\ 	ext{and}\ x_{n+2} = \dfrac{1 + x_{n+1}}{x_n},\ n = 0,1,2,3,\ldots$, then $x_{2021}$ is equal to

Accepted Answer

Find the first few terms using the recurrence.
$x_0 = 1, \quad x_1 = 2$
$x_2 = \dfrac{1+2}{1} = 3$
$x_3 = \dfrac{1+3}{2} = 2$
$x_4 = \dfrac{1+2}{3} = 1$
$x_5 = \dfrac{1+1}{2} = 1$
$x_6 = \dfrac{1+1}{1} = 2$
From $x_1$ onwards, the pattern $2, 3, 2, 1, 1$ repeats with period 5.
Since 2021 leaves rema

If x0=1,x1=2, and xn+2=1+xn+1xn, n=0,1,2,3,…x_0 = 1, x_1 = 2,\ \text{and}\ x_{n+2} = \dfrac{1 + x_{n+1}}{x_n},\ n = 0,1,2,3,\ldotsx0​=1,x1​=2, and xn+2​=xn​1+xn+1​​, n=0,1,2,3,…, then x2021x_{2021}x2021​ is equal to

If $x_0 = 1, x_1 = 2,\ \text{and}\ x_{n+2} = \dfrac{1 + x_{n+1}}{x_n},\ n = 0,1,2,3,\ldots$ , then $x_{2021}$ is equal to