Question

\begin{cases}
f(1)=15
\\\\
f(n)=f(n-1) \cdot n
\end{cases}

Answer 1

The given function \( f(n) \) is defined recursively with the base case \( f(1) = 15 \) and the recursive relation:

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

We can calculate the values of \( f(n) \) for various integers \( n \):

1. For \( n = 1 \):
\[
f(1) = 15
\]

2. For \( n = 2 \):
\[
f(2) = f(1) \cdot 2 = 15 \cdot 2 = 30
\]

3. For \( n = 3 \):
\[
f(3) = f(2) \cdot 3 = 30 \cdot 3 = 90
\]

4. For \( n = 4 \):
\[
f(4) = f(3) \cdot 4 = 90 \cdot 4 = 360
\]

5. For \( n = 5 \):
\[
f(5) = f(4) \cdot 5 = 360 \cdot 5 = 1800
\]

6. For \( n = 6 \):
\[
f(6) = f(5) \cdot 6 = 1800 \cdot 6 = 10800
\]

From these calculations, we can summarize the values obtained:

- \( f(1) = 15 \)
- \( f(2) = 30 \)
- \( f(3) = 90 \)
- \( f(4) = 360 \)
- \( f(5) = 1800 \)
- \( f(6) = 10800 \)

The pattern indicates that \( f(n) \) may relate to \( n! \) (the factorial of \( n \)) multiplied by the base value \( f(1) \). In fact, we can express \( f(n) \) as follows:

\[
f(n) = 15 \cdot n!
\]

To verify this conjecture, we can prove it by induction:

**Base Case:**
For \( n = 1 \):
\[
f(1) = 15 \cdot 1! = 15 \quad \text{(True)}
\]

**Inductive Step:**
Assume the hypothesis is true for \( n = k \):
\[
f(k) = 15 \cdot k!
\]
We need to show it is also true for \( n = k+1 \):
\[
f(k+1) = f(k) \cdot (k+1) = (15 \cdot k!) \cdot (k + 1) = 15 \cdot k! \cdot (k + 1) = 15 \cdot (k + 1)!
\]
Thus, \( f(k+1) = 15 \cdot (k+1)! \) is true.

By induction, this proves our formula holds for all \( n \geq 1 \).

In conclusion, the function can be expressed as:

\[
f(n) = 15 \cdot n!
\]