Simplifying a Geometric Series with Two Power Terms

Question

I have derived a geometric series below that I want to simplify but keep making a mess. Can anyone help? $$s = aq^{n-1}r^{0} + aq^{n-2}r^{1} + \dotsb + aq^{1}r^{n-2} + aq^{0}r^{n-1}$$

I get the following, but I know it's wrong

$$s = \frac{aq^{n-1} - (ar^{n})/q}{1 - (r/q)}$$

Answer 1

Note that \begin{align} S & = aq^{n-1}r^{0} + aq^{n-2}r^{1} + \cdots + aq^{1}r^{n-2} + aq^{0}r^{n-1}\\ & = aq^{n-1}\left(1+\frac{r}{q}+ \cdots +\frac{r^{n-1}}{q^{n-1}}\right)\\ & = aq^{n-1} \frac{1-\frac{r^n}{q^n}}{1-\frac{r}{q}} \\ & = a\frac{q^n-r^n}{q-r}. \end{align}

Answer 2

HINT

Recall that we have

$$A^n-B^n=(A-B)(A^{n-1}+A^{n-2}B+\ldots+AB^{n-2}+B^{n-1})$$

and therefore

$$A^{n-1}+A^{n-2}B+\ldots+AB^{n-2}+B^{n-1}=\frac{A^n-B^n}{A-B}=A^{n-1}\frac{1-\left(\frac B A\right)^n}{1-\frac B A}$$

Answer 3

Hint:

Use the high-school formula: $$a^n-b^n=(a-b)(a^{n-1}+a^{n-2}b+\dots+ab^{n-2}+b^{n-1}). $$

Answer 4

To find the common ratio (and prove that it is common, and hence a geometric series), divide two successive terms: $$\frac{aq^{n-1-(k+1)}r^{k+1}}{aq^{n-1-k}r^k} = \frac{r}{q}.$$ Note that this doesn't depend on $n$, so the ratio is common between all terms. The initial term is $aq^{n-1}$, and there are $n$ terms, so the sum is, $$s = aq^{n-1} \frac{\left(\frac{r}{q}\right)^n - 1}{\frac{r}{q} - 1} = aq\frac{qr^n - 1}{r - q}$$