# How to find $k$'s quotient of $\frac{-18}{(-2)^k}$

I try to solve a geometric sum, but I'm stuck. The original terms are:

$$s = 9 - \frac{9}{2} + \frac{9}{4} - \frac{9}{8} + \cdots - \frac{9}{128}$$

After some work I found that the sum would be:

$$\sum_{k=1}^8 \frac{-18}{(-2)^k}$$

My next step towards finding the sum of all terms is to use the general formula. However, this includes getting the quotient, and I don't understand how to do so in this type of scenario.

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## 1 Answer

I expect that you know that $$a+ar+ar^2+\cdots +ar^n=a\frac{r^{n+1}-1}{r-1}=a\frac{1-r^{n+1}}{1-r}$$ (if $r\ne 1$). It remains to identify $a$, $r$, and $n$.

We have $a=9$. And always we obtain the "next" term by multiplying the previous term by $-\frac{1}{2}$, so now we know that $r=-\frac{1}{2}$. Since $128=2^7$, we know that $n=7$.

Substitute in the general formula, either version, but for this case I prefer the second form, it makes the denominator positive. We have $1-r=1-\left(-\frac{1}{2}\right)=\frac{3}{2}$.

For the top, we will need to calculate $1-\left(-\frac{1}{2}\right)^8$. This is $1-\frac{1}{256}$, which you may want to simplify to $\frac{255}{256}$. Now it is just a question of putting the pieces together, and maybe simplifying somewhat.

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Thank you, I see now. But don't you mean $\frac{a(r^{n+1}-1)}{r-1}$? Or am I mistaken? – Quispiam Aug 24 '12 at 21:49
@Quispiam: Thank you for noticing! I was concentrating too much on typing the fraction correctly. Fixed. – André Nicolas Aug 24 '12 at 21:52