Vtyjkyuk

I'm solving various problems on sequences and here is one I'm stuck with:

Given $x_1 = a^1over k$ where $a > 0$, and $x_n+1 = left(aover x_nright)^1over k$ where $n in mathbb N$. Prove that:
$$
x_n = a^1-(-k)^-n over k +1
$$

Everything I've tried so far just gives me some identity. For example:

$$
beginalign
x_n+1 &= left(aover x_nright)^1over k = \
&= fraca^1over kx_n^1over k = \
&= fracx_1x_n^1over k
endalign
$$

Now try to express $x_n$:

$$
beginalign
x_n^1over k &= fracx_1x_n+1 iff \
iff x_n &= fracx_1^kx_n+1^k
endalign
$$

So this results in $x_n = x_n$ which doesn't show anything. I feel like there might be some smart substitution to translate the problem into some recurrence relation but don't see how.

Could someone please point me in the right direction?

$$x_1=a^frac1k$$
now calculate $x_2$ using $x_n+1 = left(aover x_nright)^1over k$

thus $$x_2= (fracax_1)^frac1k$$
thus $$x_2=a^frac1k*x_1^frac-1k=a^frac1k-frac1k^2$$

now solve for $x_3$ using $x_2$ obtained above

you will get $$x_3=a^frac1k-frac1k^2+frac1k^3$$
also $$x_4=a^frac1k-frac1k^2+frac1k^3-frac1k^4$$

hope you are seeing the pattern so
we can write it for $x_n$

now the problem reduces to summing the series for n(even) $$S_n=frac1k-frac1k^2+frac1k^3-frac1k^4+.....-frac1k^n$$

Hint. One may just rewrite the initial relation as
$$
left(-kright)^n+1ln (x_n+1)-left(-kright)^nln (x_n)=ln left[a^left(-kright)^nright]
$$ recognizing a telescoping sum then using logarithmic properties.