Question

Write a recursive formula for the sequence represented by the explicit formula

asubscriptn=3/4(2)^n-1

Answer 1

To create a recursive formula for the sequence given by the explicit formula

\[
a_n = \frac{3}{4} (2)^{n-1},
\]

we start by calculating the first few terms of the sequence:

- For \( n = 1 \):
\[
a_1 = \frac{3}{4} (2)^{1-1} = \frac{3}{4} (2)^0 = \frac{3}{4}.
\]

- For \( n = 2 \):
\[
a_2 = \frac{3}{4} (2)^{2-1} = \frac{3}{4} (2)^1 = \frac{3}{4} \cdot 2 = \frac{3}{2}.
\]

- For \( n = 3 \):
\[
a_3 = \frac{3}{4} (2)^{3-1} = \frac{3}{4} (2)^2 = \frac{3}{4} \cdot 4 = 3.
\]

- For \( n = 4 \):
\[
a_4 = \frac{3}{4} (2)^{4-1} = \frac{3}{4} (2)^3 = \frac{3}{4} \cdot 8 = 6.
\]

Now, let's observe how the terms relate to each other:

- From \( a_1 \) to \( a_2 \):
\[
a_2 = 2 \cdot a_1.
\]

- From \( a_2 \) to \( a_3 \):
\[
a_3 = 2 \cdot a_2.
\]

- From \( a_3 \) to \( a_4 \):
\[
a_4 = 2 \cdot a_3.
\]

We can see that the relationship is:

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

Thus, we can write the recursive formula for the sequence. The first term and the recursive relationship gives us:

\[
\begin{cases}
a_1 = \frac{3}{4}, \\
a_n = 2 \cdot a_{n-1} \quad \text{for } n > 1.
\end{cases}
\]

This forms the desired recursive formula for the sequence.