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.