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The table you provided shows the relationship between the stage \( n \) and the number of new squares \( S \). Let's examine the data and find a general rule.

From the table:

- When \( n = 1 \), \( S = 4 \)
- When \( n = 2 \), \( S = 8 \)
- When \( n = 3 \), \( S = 16 \)
- When \( n = 4 \), \( S = 32 \)
- When \( n = 5 \), \( S = 64 \)
- When \( n = 6 \), \( S = 128 \)

Looking at the pattern, you can see that the number of new squares doubles at each stage. This suggests the relationship between \( S \) and \( n \) is exponential. More specifically:

\[
S(n) = 2^n \times 4
\]

This formula represents the fact that the number of new squares at stage \( n \) is \( 4 \times 2^n \), where \( n \) starts from 1. Therefore, the general rule for \( S \) in terms of \( n \) is:

\[
S(n) = 4 \times 2^{n-1}
\]

Would you like any further clarification or examples with this formula? Here are some questions to deepen your understanding:

1. How many new squares will appear at stage \( n = 7 \) using the formula?
2. What happens to the number of new squares as \( n \) increases significantly?
3. Can you describe the significance of the base 2 in this context?
4. What does the 4 represent in the formula for \( S(n) \)?
5. How would the formula change if the initial number of squares was different?

**Tip:** When you recognize exponential growth patterns, always look for the doubling or halving factor to find a suitable exponential formula.

Learn how to derive a general rule for S in terms of n, where the number of new squares increases exponentially. The formula S(n) = 4 × 2^(n-1) is explored using a simple table of values, making it suitable for students in Grades 6-8.

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