Prove that $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$ using Mathematical induction

Question

$x_1=1$ $x_2=2$
$x_n=\frac{1}{2}(x_{n-2}+x_{n-1})$ for n $\gt$ 2. We have to prove that $|x_n-x_{n+1}|=\frac{1}{2^{n-1}}$

What I tried :
For n=1, $|x_1-x_{2}|=1 =\frac{1}{2^{0}}$
Let for n=k assumption be true. Hence $|x_k-x_{k+1}|=\frac{1}{2^{k-1}}$

For $n=k+1$
$|x_{k+1}-x_{k+2}|=\frac{1}{2}|x_{k-1}+x_{k}-x_{k}-x_{k+1}|= \frac{1}{2}|(x_{k-1}-x_{k})+(x_{k}-x_{k+1})| \le \frac{1}{2}|\frac{1}{2^{k-2}}+\frac{1}{2^{k-1}}|$

I'm stuck here, what do i do now? Can anyone help please?

Answer 1

Using triangle inequality will probably not help you, as you're supposed to prove an exact equality, not an inequality. Instead, just substitute $x_{k+2} = \frac{1}{2}(x_k + x_{k+1})$. Then, \begin{align*} |x_{k+1} - x_{k+2}| &= \left|x_{k+1} - \frac{1}{2}(x_k + x_{k+1})\right| \\ &= \left|\frac{1}{2}x_k - x_{k+1}\right| \\ &= \frac{1}{2} \cdot \frac{1}{2^{k -1}} = \frac{1}{2^k}, \end{align*} as required.

Answer 2

$$2x_n=x_{n-2}+x_{n-1}$$

$$\iff2(x_n-x_{n-1})=-(x_{n-1}-x_{n-2})$$