Question

\begin{cases}
g(1)=-29
\\\\
g(n)=g(n-1)\cdot(-4)
\end{cases}

Answer 1

The function \( g(n) \) is defined recursively with the initial condition \( g(1) = -29 \) and the recurrence relation \( g(n) = g(n-1) \cdot (-4) \). We can compute the values of \( g(n) \) for the first few integers \( n \) as follows:

1. **Calculate \( g(2) \)**:
\[
g(2) = g(1) \cdot (-4) = -29 \cdot (-4) = 116
\]

2. **Calculate \( g(3) \)**:
\[
g(3) = g(2) \cdot (-4) = 116 \cdot (-4) = -464
\]

3. **Calculate \( g(4) \)**:
\[
g(4) = g(3) \cdot (-4) = -464 \cdot (-4) = 1856
\]

4. **Calculate \( g(5) \)**:
\[
g(5) = g(4) \cdot (-4) = 1856 \cdot (-4) = -7424
\]

Now we notice a pattern. The values for \( g(n) \) seem to be alternating in sign and increasing in absolute value.

Let's derive a formula for \( g(n) \):

From our calculations, we can express \( g(n) \) in terms of \( g(1) \):
\[
g(1) = -29
\]
\[
g(2) = -29 \cdot (-4)^1 = -29 \cdot (-4)
\]
\[
g(3) = -29 \cdot (-4)^2
\]
\[
g(4) = -29 \cdot (-4)^3
\]
\[
g(5) = -29 \cdot (-4)^4
\]

Thus, we can generally state:
\[
g(n) = -29 \cdot (-4)^{n-1}
\]

This holds true for \( n \geq 1 \). We can also simplify this equation:
\[
g(n) = -29 \cdot (-1)^{n-1} \cdot 4^{n-1}
\]
This indicates that the sign of \( g(n) \) depends on the parity of \( n \).

### Summary:

The values calculated for \( g(n) \) are:
- \( g(1) = -29 \)
- \( g(2) = 116 \)
- \( g(3) = -464 \)
- \( g(4) = 1856 \)
- \( g(5) = -7424 \)

And the general formula for \( g(n) \) is:
\[
g(n) = -29 \cdot (-4)^{n-1} \quad \text{for } n \geq 1
\]
Or equivalently:
\[
g(n) = -29 \cdot (-1)^{n-1} \cdot 4^{n-1}
\]