Question

h(n)=-91\cdot\left(-\dfrac{1}{7}\right)^{\large{\,n-1}}Complete the recursive formula of
\[h(n)\].

Answer 1

To derive the recursive formula for the sequence defined by \( h(n) = -91 \cdot \left(-\frac{1}{7}\right)^{n-1} \), we start by determining the relationship between consecutive terms.

Given:

\[
h(n) = -91 \cdot \left(-\frac{1}{7}\right)^{n-1}
\]

We can express the previous term \( h(n-1) \):

\[
h(n-1) = -91 \cdot \left(-\frac{1}{7}\right)^{(n-1)-1} = -91 \cdot \left(-\frac{1}{7}\right)^{n-2}
\]

Now, we want to find a recursive formula that relates \( h(n) \) and \( h(n-1) \).

Let's find \( h(n) \) in terms of \( h(n-1) \):

Starting from our expressions:

\[
h(n) = -91 \cdot \left(-\frac{1}{7}\right)^{n-1}
\]
\[
h(n-1) = -91 \cdot \left(-\frac{1}{7}\right)^{n-2}
\]

Notice that:

\[
h(n) = \left(-\frac{1}{7}\right) \cdot h(n-1)
\]

This relationship leads to our recursive formula:

\[
h(n) = -\frac{1}{7} \cdot h(n-1)
\]

Where the base case would be \( h(1) \):

\[
h(1) = -91 \cdot \left(-\frac{1}{7}\right)^{1-1} = -91 \cdot 1 = -91
\]

Thus, we summarize the recursive formula as:

\[
h(n) = -\frac{1}{7} \cdot h(n-1) \quad \text{for } n > 1 \text{ with } h(1) = -91
\]