Proof by induction of whole numbers

Question

A sequence $X_1, X_2,\dots,X_n$ is defined by:
$X_1 = 1$ and $X_{k+1} = \dfrac{X_k}{X_k + 2}$ for $k\ge1$.
Show by using induction that $X_n = \dfrac1{2^n - 1}$ for all $n\ge1$.
So far I've showed that $n = 1$ in $X_n$ equals $1$, so that is "ok!".
Next I show for $k+1$ which gives me the equation:
$$\dfrac1{2^{k+1} -1} + \dfrac{X_k}{X_k + 2}$$ From here i struggle with the rest and gets confused by the $X$ in the equation.

Answer 1

We have $$x_n = \frac{1}{2^n - 1}$$

So for $n=k$ $$x_k = \frac{1}{2^k - 1}$$

Now using recurrence relation $$x_{k+1} = \frac{x_k}{x_k + 2}=\frac{\frac{1}{2^k - 1}}{\frac{1}{2^k - 1} + 2}=\frac{}{}{\frac{\frac{1}{2^k-1}}{\frac{1+2\cdot2^k-2}{2^k-1}}=\frac{1}{2^{k+1}-1}}$$

$$x_{k+1}=\frac{1}{2^{k+1}-1}$$

So, If our assumption that $x_n = \dfrac{1}{2^n - 1}$ is true for some integer $k$ then it's also true for $k+1$ and...