Let (1+√2)n=xn+yn √2 where xn, yn are integers, then :

Question

Let (1+√2)n=xn+yn √2 where xn, yn are integers, then : (A) xn2-2 yn2=(-1)n (B) xn+2 yn-xn+1=0 (C) xn2-2 yn2=1 (D) yn+1=xn+yn

Tardigrade Admin · Accepted Answer

We have (1+√2)n=xn+yn √2...(i) (1-√2)n=xn-yn √2....(i) From (i) and (ii) we get: (1+√2)n(1-√2)n=xn2-2 yn2 ⇒ (-1)n=xn2-2 yn2 Next, xn+1+yn+1 √2=(1+√2)n+1=(1+√2)(xn+√2 yn)=(xn+2 yn)+√2(xn+yn) Thus, xn+1=xn+2 yn or xn+1-xn-2 yn=0

Q. Let $(1 + 2)^{n} = x_{n} + y_{n} 2$ where $x_{n}, y_{n}$ are integers, then :

$x_{n}^{2} - 2 y_{n}^{2} = (- 1)^{n}$

$x_{n} + 2 y_{n} - x_{n + 1} = 0$

$x_{n}^{2} - 2 y_{n}^{2} = 1$

$y_{n + 1} = x_{n} + y_{n}$

Q. Let (1+2​)n=xn​+yn​2​ where xn​,yn​ are integers, then :

xn2​−2yn2​=(−1)n

xn​+2yn​−xn+1​=0

xn2​−2yn2​=1

yn+1​=xn​+yn​

Q. Let $(1 + 2)^{n} = x_{n} + y_{n} 2$ where $x_{n}, y_{n}$ are integers, then :

$x_{n}^{2} - 2 y_{n}^{2} = (- 1)^{n}$

$x_{n} + 2 y_{n} - x_{n + 1} = 0$

$x_{n}^{2} - 2 y_{n}^{2} = 1$

$y_{n + 1} = x_{n} + y_{n}$