We are given a sequence $a_k$ defined by the recurrence relation:

$a_{k+1} = 2a_k - 1, \quad \text{with} \quad a_1 = 2.$

Step 1: Find the 6th term of the sequence.

We can calculate the terms of the sequence iteratively using the given recurrence relation:

1. $a_1 = 2$
2. $a_2 = 2a_1 - 1 = 2 \times 2 - 1 = 3$
3. $a_3 = 2a_2 - 1 = 2 \times 3 - 1 = 5$
4. $a_4 = 2a_3 - 1 = 2 \times 5 - 1 = 9$
5. $a_5 = 2a_4 - 1 = 2 \times 9 - 1 = 17$
6. $a_6 = 2a_5 - 1 = 2 \times 17 - 1 = 33$

So, the 6th term of the sequence is $\boxed{33}$.

Step 2: Find the sum $\sum_{k=1}^{15} a_k$.

We need to calculate the sum of the first 15 terms of the sequence. Since we know the recurrence relation, we can continue calculating the terms:

• $a_7 = 2a_6 - 1 = 2 \times 33 - 1 = 65$
• $a_8 = 2a_7 - 1 = 2 \times 65 - 1 = 129$
• $a_9 = 2a_8 - 1 = 2 \times 129 - 1 = 257$
• $a_{10} = 2a_9 - 1 = 2 \times 257 - 1 = 513$
• $a_{11} = 2a_{10} - 1 = 2 \times 513 - 1 = 1025$
• $a_{12} = 2a_{11} - 1 = 2 \times 1025 - 1 = 2049$
• $a_{13} = 2a_{12} - 1 = 2 \times 2049 - 1 = 4097$
• $a_{14} = 2a_{13} - 1 = 2 \times 4097 - 1 = 8193$
• $a_{15} = 2a_{14} - 1 = 2 \times 8193 - 1 = 16385$

Now we sum the first 15 terms:

$\sum_{k=1}^{15} a_k = a_1 + a_2 + a_3 + \dots + a_{15} = 2 + 3 + 5 + 9 + 17 + 33 + 65 + 129 + 257 + 513 + 1025 + 2049 + 4097 + 8193 + 16385$

Calculating the sum:

$\sum_{k=1}^{15} a_k = 32784$

So, the value of $\sum_{k=1}^{15} a_k$ is $\boxed{32784}$.

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Would you like more details or further clarification on any step? Here are some related questions to explore:

1. How can you derive a general formula for $a_k$?
2. What happens if the recurrence relation changes slightly, e.g., if the term is $a_{k+1} = 2a_k + 1$?
3. How do recurrence relations like this relate to exponential growth?
4. Can you find the 20th term using a similar method?
5. How does the sum of the sequence change if we take the sum from $k = 1$ to $k = 10$?

Tip: When dealing with recurrence relations, you can often find patterns or closed-form formulas by analyzing the first few terms and then testing for consistency.