Question

a sequence is defined recursively by a1=4
an=3/4an-1

what is the explicit form of this sequence

an=4(3.4)n
an=4+3.4n
an=4times3.4(n-1)
an=4(3.4)n-1

Answer 1

To find the explicit form of the sequence defined by the recursive relationship \( a_1 = 4 \) and \( a_n = \frac{3}{4} a_{n-1} \), we can first compute a few terms of the sequence:

1. \( a_1 = 4 \)
2. \( a_2 = \frac{3}{4} a_1 = \frac{3}{4} \cdot 4 = 3 \)
3. \( a_3 = \frac{3}{4} a_2 = \frac{3}{4} \cdot 3 = \frac{9}{4} \)
4. \( a_4 = \frac{3}{4} a_3 = \frac{3}{4} \cdot \frac{9}{4} = \frac{27}{16} \)
5. \( a_5 = \frac{3}{4} a_4 = \frac{3}{4} \cdot \frac{27}{16} = \frac{81}{64} \)

From the recursive definition, each term is multiplied by \( \frac{3}{4} \) from the previous term. This suggests that we can express \( a_n \) in a form based on powers of \( \frac{3}{4} \).

To express this sequence explicitly, notice that:

\[
a_n = a_1 \cdot \left( \frac{3}{4} \right)^{n-1}
\]

Substituting \( a_1 = 4 \):

\[
a_n = 4 \cdot \left( \frac{3}{4} \right)^{n-1}
\]

We can rewrite \( \frac{3}{4} \) as \( 3 \cdot 4^{-1} \):

\[
\left( \frac{3}{4} \right)^{n-1} = 3^{n-1} \cdot 4^{-(n-1)} = \frac{3^{n-1}}{4^{n-1}}
\]

Thus,

\[
a_n = 4 \cdot \frac{3^{n-1}}{4^{n-1}} = 4 \cdot \frac{3^{n-1}}{4^{n-1}} = \frac{4 \cdot 3^{n-1}}{4^{n-1}} = \frac{4 \cdot 3^{n-1}}{4^{n}} = \frac{3^{n-1}}{4^{n-1}} \cdot 4
\]

This can also be simplified to:

\[
a_n = 4 \cdot 3^{n-1} \cdot 4^{-n + 1} = 4^{1-n} \cdot 3^{n-1}
\]

Finally, to express it more cleanly, we see that:

\[
a_n = 4 \cdot \left(\frac{3}{4}\right)^{n-1}
\]

Thus, with the form \( 4 \times (3/4)^{n-1} \), we relate this back to the answer choices provided.

Among the options, the one that matches is:

- **\( a_n = 4(3/4)^{n-1} \)**

This matches \( a_n = 4 \cdot 3^{n-1} \cdot 4^{-(n-1)} \).

To summarize, the explicit form of the sequence is:

\[
a_n = 4 \left( \frac{3}{4} \right)^{n-1}
\]

Which corresponds to the choice:

**\( a_n = 4 \times 3 \cdot 4^{-(n-1)} \)** or \( a_n = 4 \cdot (3/4)^{n-1} \).