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

Question

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.

Answer 1

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

Answer 2

We have that $x_{n+1}=x_n+x_{n-1}.$ Then $2x_{n+1}=2x_n+2x_{n-1}$ and $\underbrace{2x_{n+1}-1}_{y_{n+1}}=\underbrace{2x_n-1}_{y_n}+\underbrace{2x_{n-1}-1}_{y_{n-1}}+1$. Thus $y_{n+1}=y_n+y_{n-1}+1.$