https://file.aiquickdraw.com/mathbot/file/1729149391423ct48k8z6.jpg

The image asks us to evaluate the sum of a geometric series:

\[
S = \sum_{n=1}^{5} 6(2)^{n-1}
\]

This is a geometric series with:
- First term \( a = 6 \)
- Common ratio \( r = 2 \)
- Number of terms \( n = 5 \)

We can use the formula for the sum of a geometric series:

\[
S = \frac{a(1 - r^n)}{1 - r}
\]

### Step-by-step solution:

1. **First term \( a \):**
   The first term is given as \( a = 6 \).

2. **Common ratio \( r \):**
   The common ratio is \( r = 2 \).

3. **Number of terms \( n \):**
   The series runs from \( n = 1 \) to \( n = 5 \), so \( n = 5 \).

4. **Substitute into the formula:**

\[
S = \frac{6(1 - 2^5)}{1 - 2}
\]

Simplifying the terms:

\[
S = \frac{6(1 - 32)}{1 - 2} = \frac{6(-31)}{-1}
\]

\[
S = 186
\]

Thus, the value of \( S \) is:

\[
S = 186
\]

Would you like more details, or do you have any questions on the solution?

Here are 5 related questions to expand your understanding:
1. What is a geometric series, and how does it differ from an arithmetic series?
2. How would the result change if the common ratio \( r \) was negative?
3. Can this sum formula be applied to an infinite geometric series?
4. What happens to the sum if the number of terms \( n \) is much larger?
5. How can you derive the sum formula for a geometric series?

**Tip:** Always check whether a geometric series is finite or infinite, as this impacts which formula you should use.

Evaluate the geometric series: S = sum(6(2)^(n-1)) from n=1 to 5.

Learn how to evaluate the sum of a geometric series where S = sum(6(2)^(n-1)) from n=1 to 5. This problem involves using the geometric series sum formula and covers algebraic concepts, suitable for students in Grades 9-11.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation