Question

\begin{cases}c(1)=56\\\\
c(n)=c(n-1)\cdot \dfrac{1}{2}
\end{cases}What is the
\[4^{\text{th}}\] term in the sequence?

Answer 1

We start with the initial condition and recursive definition of the sequence given by:

\[
\begin{cases}
c(1) = 56 \\
c(n) = c(n-1) \cdot \frac{1}{2}
\end{cases}
\]

We want to find \( c(4) \). To do this, we will first calculate \( c(2) \), then \( c(3) \), and finally \( c(4) \).

1. **Calculating \( c(2) \)**:
\[
c(2) = c(1) \cdot \frac{1}{2} = 56 \cdot \frac{1}{2} = 28
\]

2. **Calculating \( c(3) \)**:
\[
c(3) = c(2) \cdot \frac{1}{2} = 28 \cdot \frac{1}{2} = 14
\]

3. **Calculating \( c(4) \)**:
\[
c(4) = c(3) \cdot \frac{1}{2} = 14 \cdot \frac{1}{2} = 7
\]

Thus, the \( 4^{\text{th}} \) term in the sequence is

\[
\boxed{7}
\]