Question

\begin{cases}
h(1)=-35
\\\\
h(n)=h(n-1) \cdot 2
\end{cases}

Answer 1

The given equations define a recursive function \( h(n) \).

1. Base case:
\[
h(1) = -35
\]

2. Recursive case:
\[
h(n) = h(n-1) \cdot 2
\]
for \( n > 1 \).

To find \( h(n) \) for various values of \( n \), we can compute the first few terms:

- For \( n = 2 \):
\[
h(2) = h(1) \cdot 2 = -35 \cdot 2 = -70
\]

- For \( n = 3 \):
\[
h(3) = h(2) \cdot 2 = -70 \cdot 2 = -140
\]

- For \( n = 4 \):
\[
h(4) = h(3) \cdot 2 = -140 \cdot 2 = -280
\]

- For \( n = 5 \):
\[
h(5) = h(4) \cdot 2 = -280 \cdot 2 = -560
\]

From this pattern, we can see that the function doubles the value of \( h \) for each increment in \( n \).

Notice the pattern can be generalized as follows:
- The value of \( h(n) \) can be expressed in terms of \( h(1) \):
\[
h(n) = h(1) \cdot 2^{n-1}
\]

Substituting \( h(1) = -35 \) gives:
\[
h(n) = -35 \cdot 2^{n-1}
\]

Thus, the explicit formula for \( h(n) \) is:
\[
h(n) = -35 \cdot 2^{n-1}
\]

You can further verify a few values with this formula:
- For \( n = 2 \): \( h(2) = -35 \cdot 2^{1} = -70 \)
- For \( n = 3 \): \( h(3) = -35 \cdot 2^{2} = -140 \)
- For \( n = 4 \): \( h(4) = -35 \cdot 2^{3} = -280 \)
- For \( n = 5 \): \( h(5) = -35 \cdot 2^{4} = -560 \)

All calculations match the values obtained recursively.