Can someone please help me with this problem? Thanks!

Evaluate the infinite geometric series:

8+4+2+1+...

I thought you needed to know the number of terms to evaluate, but how do you know them if it is infinite?

The limit of Sn = 1 + r + r^2 + r^3 + r^4 + = 1/(1 - r) -1 < r < 1

Therefore, the sum of the series 8, 4, 2, 1, 1/2, 1/4, 1/8, ....becomes 8 + 4 + 2 + 1/(1 - 1/2) = 16

I'm not quite sure I understand your formula. The formula I received for geometric series is Sn=a1 (1-r^n)/(1-r) where a1 is the first term and r is what you multiply by and n is given or is the number of terms.

Is the -1 < r < 1 the same as what we got: |r| < 1, which means it is convergent and |r| > or = 1 so it is divergent?

While it is a geometric series, it is a converging series with an infinite number of terms.

The limit, or number that the series converger on is 1/( - r) where -1<r<+1.

Applying Sg = a(r^n - 1)/(r - 1) yields

Sg = 8[(1/2)^inf. - 1]/(1/2 - 1) =

8[0 - 1]/(-1/2) = (-8)/(-1/2) = 16.

I am really confused.

This formula you gave: a(r^n - 1)/(r - 1) is very similar to mine: a1 (1-r^n)/(1-r).

Is your way the only way to do it? Do you know how to do it the way my teacher taught?

No, there are multiple ways to evaluate an infinite geometric series. The formula you mentioned, Sn = a1 (1-r^n)/(1-r), is another valid formula to use. Let's go through the steps using your formula to evaluate the series.

The given series is 8 + 4 + 2 + 1 + ...

To use the formula Sn = a1 (1-r^n)/(1-r), we need to find the first term (a1) and the common ratio (r).

In this series, the first term is 8 (a1 = 8), and the common ratio is 1/2 (r = 1/2), since each term is half of the previous term.

Now let's substitute these values into the formula and simplify:

Sn = 8 (1 - (1/2)^n) / (1 - 1/2)

As n approaches infinity, the term (1/2)^n approaches 0. Therefore, we have:

Sn = 8 (1 - 0) / (1 - 1/2)
= 8 (1/2)
= 4

So, using your formula, the sum of the given infinite geometric series is 4.

Both the formulas - Sn = 1 + r + r^2 + r^3 + r^4 + ... and Sn = a1 (1-r^n)/(1-r) - are valid and can be used to evaluate infinite geometric series. It's just a matter of using the appropriate formula based on the given information and what you are comfortable with.

I hope this clears up your confusion! Let me know if you have any further questions.