g(n)=25-49(n-1)

To complete the recursive formula for \( g(n) \), we start from the given function:

\[
g(n) = 25 - 49(n - 1)
\]

First, we will calculate \( g(1) \) to find the base case of the recursion:

\[
g(1) = 25 - 49(1 - 1) = 25 - 49(0) = 25 - 0 = 25
\]

Now, we need to express \( g(n) \) in terms of \( g(n-1) \). Let's compute \( g(n-1) \):

\[
g(n-1) = 25 - 49((n-1) - 1) = 25 - 49(n - 2) = 25 - 49n + 98 = 123 - 49n
\]

Now we will rewrite \( g(n) \) in terms of \( g(n-1) \):

\[
g(n) = 25 - 49(n - 1) = 25 - 49n + 49 = 74 - 49n
\]

Next, we express the difference \( g(n) - g(n-1) \):

\[
g(n) - g(n-1) = (74 - 49n) - (123 - 49n) = 74 - 49n - 123 + 49n = 74 - 123 = -49
\]

So we can establish the recursive relationship:

\[
g(n) = g(n-1) - 49
\]

Putting everything together, we have:

\[
\begin{align*}
g(1) &= 25 \\
g(n) &= g(n-1) - 49 \quad \text{for } n > 1
\end{align*}
\]

Thus, the completed recursive formula is:

\[
g(1) = 25
\]
\[
g(n) = g(n-1) - 49 \quad \text{for } n > 1
\]