Question

a_100 = 2 × a_100-1
Write an exponential function to represent your geometric sequence. Show that
it has the same structure as your explicit formula

Answer 1

To represent your geometric sequence using an exponential function, we need to recognize the pattern and the common ratio in the sequence. You have provided a recurrence relation for the sequence:

\[
a_n = 2 \times a_{n-1}
\]

This indicates that each term is obtained by multiplying the previous term by 2. The initial term \( a_1 \) will be the starting point of the sequence.

Let's express the first few terms of the sequence to understand the pattern:

- Assume \( a_1 = a \) (this is the first term).
- Then, \( a_2 = 2 \times a_1 = 2a \)
- Next, \( a_3 = 2 \times a_2 = 2 \times (2a) = 4a \)
- And \( a_4 = 2 \times a_3 = 2 \times (4a) = 8a \)

From this, we can generalize the sequence:

\[
a_n = 2^{n-1} \times a_1
\]

If we assume \( a_1 = a \), we can rewrite this as:

\[
a_n = a \cdot 2^{n-1}
\]

This is the explicit formula for your geometric sequence where \( a \) is the first term and 2 is the common ratio.

Now, to connect this to the exponential function:

The exponential function can be represented in the same structure as:

\[
f(n) = a \cdot r^{n-1}
\]

where \( r \) is the common ratio. In your case, this translates to:

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

Thus, this matches with the explicit formula we derived. We can see that both forms have the same structure. You have an initial term multiplied by a constant raised to the power of \( n-1 \), which represents the growth of the sequence in both cases. Therefore, we can conclude that:

\[
a_n = a \cdot 2^{n-1}
\]
is the explicit formula, and it has the same structure as the exponential function representing the geometric sequence.