Q:

If $y_n = 2x_n-1$, show that $y_{n+1} = y_n + y_{n-1} + 1$

If $y_n = 2x_n-1$, how do you show  $y_{n+1} = y_n + y_{n-1} + 1$
with $y_0 = 1$ and $y_1 = 1$?
Would you start with $y_{n+1} = y_n + y_{n-1} + 1$, find a formula for $y_n$ and then compare it with $2x_n-1$?
Note: $(x_n)$ is the ordinary Fibonacci sequence.

A:

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

