Question

\begin{cases}
f(1)=72
\\\\
f(n)=f(n-1)+9
\end{cases}

Answer 1

The given function \( f(n) \) is defined as follows:

\[
f(1) = 72
\]
\[
f(n) = f(n-1) + 9 \quad \text{for } n > 1
\]

This recursive definition indicates that each term in the sequence increases by 9 from the previous term. We can find the general expression for \( f(n) \) using the recurrence relation.

Let's calculate the first few values of \( f(n) \):

- For \( n = 2 \):
\[
f(2) = f(1) + 9 = 72 + 9 = 81
\]

- For \( n = 3 \):
\[
f(3) = f(2) + 9 = 81 + 9 = 90
\]

- For \( n = 4 \):
\[
f(4) = f(3) + 9 = 90 + 9 = 99
\]

- For \( n = 5 \):
\[
f(5) = f(4) + 9 = 99 + 9 = 108
\]

From this, we can observe a pattern. Specifically, it appears that \( f(n) \) can be expressed as a linear function of \( n \).

In general, the formula for \( f(n) \) can be written as:

\[
f(n) = f(1) + (n - 1) \cdot 9
\]

Substituting \( f(1) = 72 \):

\[
f(n) = 72 + (n - 1) \cdot 9
\]

We can simplify this expression:

\[
f(n) = 72 + 9n - 9
\]
\[
f(n) = 9n + 63
\]

Thus, the closed-form expression for \( f(n) \) is:

\[
\boxed{f(n) = 9n + 63}
\]