Basic Geometric Series this is very easy I think but I don't have much experience with geometric series. I need to find/simplify the sum

$x^k+x^{k+1}+\dots+x^{k+m}$ 

where $x\in\mathbb{R}, k\in\mathbb{Z},m\in\mathbb{N}$

So I'm thinking I just write this as

$x^k+x\times x^k+x^2\times x^k+\dots+x^m\times x^k$

Now isn't this a geometric series with $a=x^k$ and $r=x$? So it should have the sum

$$s=\frac{x^k(1-x^{k+m})}{1-x}=\frac{x^k-x^{2k+m}}{1-x}$$

Is this the correct answer?
$$S=x^k+x^{k+1}+\dots+x^{k+m}$$
No need to know by heart some formula. Just multiply by $(1-x)$ :
$$S(1-x)=(x^k+x^{k+1}+\dots+x^{k+m})(1-x)\\=x^k+x^{k+1}+x^{k+2}+\dots+x^{k+m}-x^{k+1}-x^{k+2}-\dots-x^{k+m}-x^{k+m+1}$$
Simplify (this is a telescoping series) :
$$S(1-x)=x^k-x^{k+m+1}$$
$$S=\frac{x^k-x^{k+m+1}}{1-x}$$
The geometric serie  (( 1+x+x^2........+x^{m+1}) has nothing to do with k it doesn't depend on k. It depends on m rather.

The correct term should be $x^{m+1}$ and not $x^{m+k}$, so the geometric serie has the sum:

$$S_m= \frac {X^{m+1} -1}{X-1}$$

And then you have to multiply by $x^k$ to get the whole sum.