https://file.aiquickdraw.com/mathbot/file/1730828104444qcsjpgw5.jpg

The expression in the image is a summation:

\[
\sum_{n=1}^{3} \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{n-1}
\]

To evaluate this, let’s break down the terms:

1. The summation runs from \( n = 1 \) to \( n = 3 \).
2. The general term is \( \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{n-1} \).
3. We can simplify each term inside the summation as:
   \[
   \left(\frac{1}{2}\right) \cdot \left(\frac{1}{2}\right)^{n-1} = \frac{1}{2^n}
   \]

Thus, the summation becomes:

\[
\sum_{n=1}^{3} \frac{1}{2^n}
\]

Now, we can calculate each term individually:

- For \( n = 1 \): \( \frac{1}{2^1} = \frac{1}{2} \)
- For \( n = 2 \): \( \frac{1}{2^2} = \frac{1}{4} \)
- For \( n = 3 \): \( \frac{1}{2^3} = \frac{1}{8} \)

Now, add these values together:

\[
\frac{1}{2} + \frac{1}{4} + \frac{1}{8} = \frac{4}{8} + \frac{2}{8} + \frac{1}{8} = \frac{7}{8}
\]

Therefore, the result of the summation is:

\[
\frac{7}{8}
\]

Would you like further details or have any questions?

Here are some additional questions for further exploration:
1. What happens if we extend this summation to infinity?
2. How would the result change if the initial term were different, say \( \frac{1}{3} \)?
3. How can this summation be represented as a geometric series?
4. What is the formula for the sum of a finite geometric series?
5. How would the sum change if the base in each term were different, such as \( \frac{1}{3} \) instead of \( \frac{1}{2} \)?

**Tip:** When working with summations involving powers, look for patterns like geometric progressions to simplify the calculation.

Evaluate the summation \( \sum_{n=1}^{3} \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right)^{n-1} \).

This solution demonstrates how to evaluate the summation \( \sum_{n=1}^{3} \left( \frac{1}{2} \right) \cdot \left( \frac{1}{2} \right)^{n-1} \) using concepts of geometric series and powers. The problem is broken down into a detailed step-by-step approach to find the sum as \( \frac{7}{8} \). Suitable for students in Grades 9-12.

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