Question

g(n)=-72\cdot\left(\dfrac{1}{6}\right)^{\large{\,n-1}}

Answer 1

The function you have provided is \( g(n) = -72 \cdot \left( \frac{1}{6} \right)^{n-1} \).

This function can be analyzed in various ways. Here are some aspects:

1. **General Form**: This is an exponential decay function because the base of the exponent, \(\frac{1}{6}\), is between 0 and 1. The presence of the negative sign means that the values of the function will always be negative as \( n \) increases.

2. **Behavior**:
- As \( n \) increases, \( \left( \frac{1}{6} \right)^{n-1} \) gets smaller and smaller (approaching 0).
- Consequently, \( g(n) \) will also approach 0, but will remain negative.

3. **Values**:
You can compute specific values of \( g(n) \):
- For \( n = 1 \):
\[
g(1) = -72 \cdot \left( \frac{1}{6} \right)^{1-1} = -72
\]
- For \( n = 2 \):
\[
g(2) = -72 \cdot \left( \frac{1}{6} \right)^{2-1} = -72 \cdot \frac{1}{6} = -12
\]
- For \( n = 3 \):
\[
g(3) = -72 \cdot \left( \frac{1}{6} \right)^{3-1} = -72 \cdot \frac{1}{36} = -2
\]
- For \( n = 4 \):
\[
g(4) = -72 \cdot \left( \frac{1}{6} \right)^{4-1} = -72 \cdot \frac{1}{216} = -\frac{1}{3}
\]

4. **Limit**:
\[
\lim_{n \to \infty} g(n) = 0
\]

5. **Graph Representation**:
The graph of this function shows a decreasing exponential curve starting at \(-72\) for \( n = 1\) and approaching \( 0\) as \( n \) becomes large.

If you would like to know more about a particular aspect about this function, feel free to ask!