# Thread: have difficulties with Geometric series questions

1. ## have difficulties with Geometric series questions

Question:

The problem I have is with the worked solutions is how did the author come up with:
$\displaystyle \frac{1}{a}(\frac{r^{n-1}+r^{n-1}+r^{n-2}+r^{n-31}+...+r^{0}}{r^{(n-1)(n-2)(n-3)..3.2.1}})$

I thought that $\displaystyle \frac{1}{a}+\frac{1}{ar} + ....+ \frac{1}{ar^{n-1}} = \frac{1}{a}(r^{0}+r^{-1}+r^{-2}+ r^{-3})$

therefore r = common ratio = r^{-1}

and lastly how did the author need up with $\displaystyle \frac{a^{n-1}n(n-1)}{2P_{n}}$ where $\displaystyle P_{n} = a^{n}\frac{r^{n(n-1)/2}}{2}$ as the answer

I thought $\displaystyle P_{n} = a^{n}\frac{r^{n(n-1)/2}}{2}$ is wrong.
Shouldn't $\displaystyle P_{n} = a^{n}r^{\frac{n(n-1)}{2}}$ ?

Please if someone can be kind to explain to me thank you

$\displaystyle \frac{1}{(a)(ar).....(ar^{n-1})}$ I did this question

2. ## Re: have difficulties with Geometric series questions

\begin{align*} &\phantom{=}\displaystyle \sum_{k=0}^{n-1} \dfrac{1}{a r^k} \\ \\ &=\dfrac 1 a \sum_{k=0}^{n-1}r^{-k}\\ \\ &= \dfrac 1 a \dfrac{r^{-n}-1}{r^{-1}-1}\\ \\ &=\dfrac 1 a \dfrac{r^{1-n} - r}{1-r} \\ \\ &=\dfrac{r^{1-n}}{a}\dfrac{r^n-1}{r-1} \end{align*}

\begin{align*} \displaystyle S_n &= a \sum_{k=0}^{n-1}r^k \\ \\ &= a \dfrac{r^n-1}{r-1} \end{align*}

so

$\displaystyle \sum_{k=0}^{n-1} \dfrac{1}{a r^k} = \dfrac{r^{1-n}S_n}{a^2}$

3. ## Re: have difficulties with Geometric series questions

Originally Posted by romsek
\$\
Thank you very much for taking your time to help me with this question