May I know why is that so? Because I was trying to prove $(N+\sqrt{N^2-1})^k$, where k is a positive integer, differs from the integer nearest to it by less than $(2N-\frac{1}{2})^{-k}$. I checked the answer, and it said that $(N+\sqrt{N^2-1})^k + (N+\sqrt{N^2-1})^{-k}$ is the solution to the recurrence sequence, with $x_0 =2, x_1 =2N$. Hence, it is an integer. So$(N+\sqrt{N^2-1})>2N-\frac{1}{2} $.
Could you guys pls explain why is that so to me?
Thank you very much for your reply
The claim in the title is false: $(N+\sqrt{N^2-1})^k + (N+\sqrt{N^2-1})^{-k}$ satisfies $x_{k+2}=2N x_{k+1} - x_k$.
In general, $x_k=a^k+b^k$ satisfies $x_{k+2}=(a+b) x_{k+1} - (ab) x_k$.