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

To express the function \( g(n) = 25 - 49(n-1) \) in a recursive format, we first need to express it in terms of \( g(n-1) \).

Starting from the given equation:

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

Let's simplify this equation:

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

Now, let's find \( g(n-1) \):

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

Now we have two consecutive terms:
1. \( g(n) = 74 - 49n \)
2. \( g(n-1) = 123 - 49n \)

Next, we want to express \( g(n) \) in terms of \( g(n-1) \):

From \( g(n-1) \):

\[
g(n-1) = 123 - 49n
\]
Rearranging this gives:

\[
49n = 123 - g(n-1)
\]
And hence:

\[
g(n) = 74 - (123 - g(n-1))
\]
This simplifies to:

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

Thus, the recursive formula can be given as:

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

And we need a base case, which can be calculated by setting \( n = 1 \):

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

Thus, the complete recursive definition is:

\[
g(n) = g(n-1) - 49 \quad \text{for } n > 1
\]
with the base case:

\[
g(1) = 25
\]