Question

Let $(1+\sqrt{2})^n=x_n+y_n \sqrt{2}$ where $x_n, y_n$ are integers, then :

• A. $x_n^2-2 y_n^2=(-1)^n$
• B. $x_n+2 y_n-x_{n+1}=0$
• C. $x_n^2-2 y_n^2=1$
• D. $y_{n+1}=x_n+y_n$

Answer 1

Answer: (A), (B), (D)

We have $(1+\sqrt{2})^n=x_n+y_n \sqrt{2}$...(i)
$(1-\sqrt{2})^n=x_n-y_n \sqrt{2}$....(i)
From (i) and (ii) we get :
$(1+\sqrt{2})^n(1-\sqrt{2})^n=x_n^2-2 y_n^2 \Rightarrow (-1)^n=x_n^2-2 y_n^2$
Next, $ x_{n+1}+y_{n+1} \sqrt{2}=(1+\sqrt{2})^{n+1}=(1+\sqrt{2})\left(x_n+\sqrt{2} y_n\right)=\left(x_n+2 y_n\right)+\sqrt{2}\left(x_n+y_n\right)$
Thus, $ x_{n+1}=x_n+2 y_n$ or $x_{n+1}-x_n-2 y_n=0$