Examine the sequence.

4, 8, 16, 32, ...

Which function represents the sequence?

	
f(n)=4n;  n=1, 2, 3, −−−−−

	
f(n)=4n−1;  n=1, 2, 3, −−−−−

	
f(n)=4(2n);  n=1, 2, 3, −−−−−

	
f(n)=4(2n−1);  n=1, 2, 3, −−−−−

To identify which function represents the given sequence \( 4, 8, 16, 32, \dots \), let's analyze the pattern:

Each term in the sequence is obtained by multiplying the previous term by 2. This indicates an **exponential growth** with a base of 2.

1. The first term \( f(1) = 4 \).
2. The second term \( f(2) = 8 \).
3. The third term \( f(3) = 16 \).
4. The fourth term \( f(4) = 32 \).

This sequence can be described as \( f(n) = 4 \cdot 2^{n-1} \), where \( n \) represents the position in the sequence.

Now, let's match this with the options:

1. **\( f(n) = 4n \)**: This represents a linear function, but our sequence is not linear. It grows exponentially, not linearly, so this is incorrect.
   
2. **\( f(n) = 4n - 1 \)**: This is still a linear function with a subtraction, which doesn't match the exponential growth of the sequence, so this is incorrect.

3. **\( f(n) = 4(2^n) \)**: This represents an exponential function, but it starts from \( 4 \cdot 2^1 = 8 \) when \( n = 1 \), which is not correct for the first term of the sequence. This is incorrect.

4. **\( f(n) = 4(2^{n-1}) \)**: This function starts with \( 4 \cdot 2^{1-1} = 4 \cdot 1 = 4 \), which matches the first term of the sequence. When \( n = 2 \), \( f(2) = 4 \cdot 2^{2-1} = 4 \cdot 2 = 8 \), and so on. This correctly represents the sequence.

Therefore, the correct function is:

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

Would you like further details or clarifications? Here are 5 related questions:

1. How would the function change if the first term of the sequence was 5 instead of 4?
2. Can you describe the difference between linear and exponential growth in sequences?
3. How would you generalize a function for sequences of powers of 3?
4. How can this sequence be represented recursively instead of with an explicit function?
5. What happens to the function if we start from a different index, say \( n = 0 \)?

**Tip:** When solving sequence problems, always check the nature of the growth—whether it's linear, quadratic, or exponential—to narrow down the correct formula.

Examine the sequence. 4, 8, 16, 32, ... Which function represents the sequence?
f(n)=4n;  n=1, 2, 3, −−−−−
f(n)=4n−1;  n=1, 2, 3, −−−−−
f(n)=4(2n);  n=1, 2, 3, −−−−−
f(n)=4(2n−1);  n=1, 2, 3, −−−−−

Explore the solution to the sequence 4, 8, 16, 32 by identifying the correct exponential function. Learn the underlying algebraic concepts, including the formula f(n) = 4(2^(n-1)) and the exponential growth pattern. Suitable for students in Grades 8-10.

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