Question 22.

If x0x_{0} = 1, x1x_{1} = 2, and xn+2x_{n + 2} = 1+xn+1xn\frac{1+x_{n+1}}{x_{n}} , n = 0, 1, 2, 3,..., then x2021x_{2021} is equal to?

A
4
B
3
C
1
D
2
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Question Explanation

Text Explanation

x0=1x_0 = 1

x1=2x_1 = 2

x2=1+x1x0=1+21=3x_2 = \frac{1 + x_1}{x_0} = \frac{1 + 2}{1} = 3

x3=1+x2x1=1+32=2x_3 = \frac{1 + x_2}{x_1} = \frac{1 + 3}{2} = 2

x4=1+x3x2=1+23=1x_4 = \frac{1 + x_3}{x_2} = \frac{1 + 2}{3} = 1

x5=1+x4x3=1+12=1x_5 = \frac{1 + x_4}{x_3} = \frac{1 + 1}{2} = 1

x6=1+x5x4=1+11=2x_6 = \frac{1 + x_5}{x_4} = \frac{1 + 1}{1} = 2

Hence, the series begins to repeat itself after every 5 terms.  

Terms whose number is of the form 5n5n are 1, 5n+15n+1 are 2, and so on, where n=0,1,2,3,n = 0,1,2,3,\dots

2021 is of the form 5n+15n + 1.  

Hence, its value will be 2.


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