Question

Walk-through needed

Answer 1

@AnswerMyQuestions @iGreen @ganeshie8

Answer 2

You could add these sums by adding the terms one-by-one, but I think it's worth going through the formula for the geometric series.

The formula is:
\[\sum_{i=1}^{n} ar^{n-1}=a\left(\frac{1-r^n}{1-r} \right)\]

So for example, let's say we had the sum:
\[\sum_{i=1}^{8}5(4)^{i-1} \], let's look how we apply the formula one-by-one:
Here, we notice that the 8 above the \(\sum\) gives us the \(n\) in the formula.
So far:
\[ \sum_{i=1}^{\color{red}{8}}5(4)^{i-1}=a\left(\frac{1-r^{\color{red}{8}}}{1-r} \right)\ \]
Now we identify what \(a\) and \(r\) are. We compare the factors to the original formula:
\( \color{blue}{a}\color{orange}{r}^{i-1}=\color{blue}{5}(\color{orange}{4})^{i-1}\)

This tells us that \(\color{blue}{a=5}\) and \(\color{orange}{r=4}\), so... our formula becomes:

\[ \sum_{i=1}^{\color{red}{8}}\color{blue}{5}(\color{orange}{4})^{i-1}=\color{blue}{5}\left(\frac{1-\color{orange}{4}^{\color{red}{8}}}{1-\color{orange}{4}} \right)\ \]

Now it's just a matter of using a calculator to find the result. \[5\left(\frac{1-4^8}{1-4}\right) =5\left(\frac{-65,353}{-3}\right)=5(21,845)=109,225\]

Answer 3

Also, if you just have something that looks like it has the form \[ \sum_{i=1}^nr^{i-1}\] this just means that \(a=1\)