Question

write a recursive rule for the sequnce: f(n)=1990x(1/2)^n^-1

Answer 1

To write a recursive rule for the sequence defined by the formula \( f(n) = 1990 \cdot \left( \frac{1}{2} \right)^{n-1} \), we first need to express it in a form that relates \( f(n) \) to \( f(n-1) \).

From the given formula, we can observe how \( f(n) \) relates to \( f(n-1) \):

\[
f(n-1) = 1990 \cdot \left( \frac{1}{2} \right)^{(n-1)-1} = 1990 \cdot \left( \frac{1}{2} \right)^{n-2}
\]

Now consider \( f(n) \):

\[
f(n) = 1990 \cdot \left( \frac{1}{2} \right)^{n-1} = \frac{1}{2} \cdot \left( 1990 \cdot \left( \frac{1}{2} \right)^{n-2} \right) = \frac{1}{2} \cdot f(n-1)
\]

Thus, the recursive relation can be expressed as:

\[
f(n) = \frac{1}{2} \cdot f(n-1) \quad \text{for } n \geq 2
\]

Additionally, we need a base case to complete the recursive definition. We can compute the first term:

\[
f(1) = 1990 \cdot \left( \frac{1}{2} \right)^{1-1} = 1990 \cdot 1 = 1990
\]

Putting it all together, the recursive definition of the sequence is:

\[
\begin{align*}
f(1) & = 1990, \\
f(n) & = \frac{1}{2} \cdot f(n-1) \quad \text{for } n \geq 2.
\end{align*}
\]