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

Complete the recursive formula of \[g(n)\]

1 answer

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:

\[ g(1) = 25 \]