The summation \(\sum_{k=1}^8 (2k-1)\) represents the sum of the first 8 odd numbers. Let's compute it step by step:

### Step-by-Step Solution:
\[
\sum_{k=1}^8 (2k-1) = (2 \cdot 1 - 1) + (2 \cdot 2 - 1) + (2 \cdot 3 - 1) + (2 \cdot 4 - 1) + \ldots + (2 \cdot 8 - 1)
\]

Expanding this, we get:
\[
= 1 + 3 + 5 + 7 + 9 + 11 + 13 + 15
\]

Now, summing these values:
\[
1 + 3 + 5 + 7 + 9 + 11 + 13 + 15 = 64
\]

### Final Answer:
\[
\sum_{k=1}^8 (2k-1) = 64
\]

Would you like further explanation or have any other questions?

Here are 5 related questions:
1. What is the sum of the first 10 odd numbers?
2. Can you generalize a formula to sum the first \(n\) odd numbers?
3. How would the summation change if the upper limit were \(k=20\)?
4. What if the expression inside the summation were \((2k + 1)\) instead of \((2k - 1)\)?
5. What is the average of the first 8 odd numbers?

**Tip**: The sum of the first \(n\) odd numbers is always \(n^2\).

Explore the solution to the summation ∑_{k=1}^8 (2k-1), which involves calculating the sum of the first 8 odd numbers. Learn how the summation results in 64, along with key concepts in arithmetic progression and summation formulas. Suitable for students in Grades 6-8.

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