Let's solve the summation problem step by step:

The given problem is:

$\sum_{n=1}^6 \left(\frac{1}{2}n + 2\right)$

Step 1: Expand the summation

The general formula for the summation is: $\sum_{n=1}^6 \left(\frac{1}{2}n + 2\right) = \sum_{n=1}^6 \frac{1}{2}n + \sum_{n=1}^6 2$

Step 2: Solve each component

1. For $\sum_{n=1}^6 \frac{1}{2}n$: The summation of $n$ from 1 to $6$ is given by: $\sum_{n=1}^6 n = 1 + 2 + 3 + 4 + 5 + 6 = \frac{6(6+1)}{2} = 21$ So: $\sum_{n=1}^6 \frac{1}{2}n = \frac{1}{2} \cdot 21 = 10.5$

2. For $\sum_{n=1}^6 2$: Since $2$ is a constant, the summation is: $\sum_{n=1}^6 2 = 2 \cdot 6 = 12$

Step 3: Combine the results

Add the two results together: $\sum_{n=1}^6 \left(\frac{1}{2}n + 2\right) = 10.5 + 12 = 22.5$

Final Answer:

$\boxed{22.5}$

Let me know if you'd like further details or have any questions!

Related Questions:

1. How do we derive the formula for the summation of $n$ from 1 to $k$?
2. Can this problem be solved using properties of linearity in summation?
3. What is the role of constant summations like $\sum c$ and how do we evaluate them?
4. How does this summation change if the upper limit changes (e.g., $n=10$ instead of $6$)?
5. What happens if the term $\frac{1}{2}n$ has a different coefficient?

Tip:

Breaking a summation into smaller components, as shown here, helps to simplify the calculation significantly!