Basic Geometric Series

Question

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?

the geometric serie dosent depend on k it's $x^{m+1}$ and not $x^{m+k}$ — user577215664, Oct 16 '17 at 07:50

score 5 · Accepted Answer · answered Oct 16 '17 at 08:20

$$S=x^k+x^{k+1}+\dots+x^{k+m}$$ 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) : $$S(1-x)=x^k-x^{k+m+1}$$ $$S=\frac{x^k-x^{k+m+1}}{1-x}$$

score 2 · Answer 2 · answered Oct 16 '17 at 07:50

2

No, that is not the correct answer. The correct answer is$$\frac{x^k-x^{k+m+1}}{1-x}.$$

answered Oct 16 '17 at 07:50

José Carlos Santos

427,504

Why is the exponent (k+m+1) and not (k+m+k)? – Corey Oct 16 '17 at 08:02
@Corey Because the formula for the sum of $a+ar+ar^2+ar^3+\cdots+ar^n$ is$$\frac{a-ar^{n+1}}{1-r}.$$In the present situation, $a=x^k$ and $r=x$. – José Carlos Santos Oct 16 '17 at 08:16

score 0 · Answer 3 · answered Oct 16 '17 at 08:05

0

Since $$1+x+\ldots+x^m=\frac{x^{m+1}-1}{x-1}$$ Then $$x^k(1+x+\ldots+x^m)=\frac{x^{m+1+k}-x^k}{x-1}$$

answered Oct 16 '17 at 08:05

ajotatxe

65,084

Basic Geometric Series

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