Partial sum of Geometric sequence

Question

Can someone explain to me how the following summation goes from the left to the right:

$\sum_{i=1}^{r}n(1-p)^{i-1} = \frac{n(1-(1-p)^r)}{p}$.

I have used the formula for a Geometric series. My common ratio is: $(1-p)$. I get the following:

$\sum_{i=1}^{r}n(1-p)^{i-1} = n\cdot\frac{1-(1-p^r)}{1-(1-p)}$.

From here, I simply don't know how to get to $\frac{n(1-(1-p)^r)}{p}$

Answer 1

In your question, $n$ is a constant, which has nothing to do with $i$. The first implication of what you have done is incorrect.

Note that $n(1-p)^{i-1}$ means $n\cdot ((1-p)^{i-1})$, and

$$ \sum_{i=1}^{r}n(1-p)^{i-1}=n\sum_{i=1}^{r}(1-p)^{i-1} $$

all you need to work out is to find a formula for $$ \sum_{i=1}^{r}x^{i-1} $$ and then set $x=1-p$.

Answer 2

\begin{align*} \sum_{i=1}^{r}n(1-p)^{i-1} & =n\sum_{i=0}^{r-1}(1-p)^i =[\text{\(r\) terms}] =n\cdot\frac{1-(1-p)^r}{1-(1-p)} =\frac{n\bigl(1-(1-p)^r\bigr)}{p} \end{align*}