Question

\begin{cases}
f(1)=25
\\\\
f(n)=f(n-1) \cdot \left(-\dfrac{1}{5}\right)
\end{cases}

Answer 1

To analyze the recursive function defined in the given equations, we start with the given base case and the recursive definition:

1. Base case: \( f(1) = 25 \)
2. Recursive case: \( f(n) = f(n-1) \cdot \left(-\frac{1}{5}\right) \)

Now let's compute the values of \( f(n) \) for the first few integers:

- For \( n = 2 \):
\[
f(2) = f(1) \cdot \left(-\frac{1}{5}\right) = 25 \cdot \left(-\frac{1}{5}\right) = -5
\]

- For \( n = 3 \):
\[
f(3) = f(2) \cdot \left(-\frac{1}{5}\right) = -5 \cdot \left(-\frac{1}{5}\right) = 1
\]

- For \( n = 4 \):
\[
f(4) = f(3) \cdot \left(-\frac{1}{5}\right) = 1 \cdot \left(-\frac{1}{5}\right) = -\frac{1}{5}
\]

- For \( n = 5 \):
\[
f(5) = f(4) \cdot \left(-\frac{1}{5}\right) = -\frac{1}{5} \cdot \left(-\frac{1}{5}\right) = \frac{1}{25}
\]

- For \( n = 6 \):
\[
f(6) = f(5) \cdot \left(-\frac{1}{5}\right) = \frac{1}{25} \cdot \left(-\frac{1}{5}\right) = -\frac{1}{125}
\]

From the calculations, we can observe a pattern. It seems that at each step the value is:

- Changing sign
- Being multiplied by a factor of \( \frac{1}{5} \)

Now let's find a general formula for \( f(n) \).

Noticing \( f(1) = 25 \), we can express the function in terms of powers of \( -\frac{1}{5} \):

\[
f(n) = 25 \cdot \left(-\frac{1}{5}\right)^{n-1}
\]

Let's verify this formula by plugging in values:

- For \( n=1 \):
\[
f(1) = 25 \cdot \left(-\frac{1}{5}\right)^{1-1} = 25 \cdot 1 = 25
\]

- For \( n=2 \):
\[
f(2) = 25 \cdot \left(-\frac{1}{5}\right)^{2-1} = 25 \cdot \left(-\frac{1}{5}\right) = -5
\]

- For \( n=3 \):
\[
f(3) = 25 \cdot \left(-\frac{1}{5}\right)^{3-1} = 25 \cdot \left(-\frac{1}{5}\right)^{2} = 25 \cdot \frac{1}{25} = 1
\]

- For \( n=4 \):
\[
f(4) = 25 \cdot \left(-\frac{1}{5}\right)^{4-1} = 25 \cdot \left(-\frac{1}{5}\right)^{3} = 25 \cdot \left(-\frac{1}{125}\right) = -\frac{1}{5}
\]

Thus, the general expression for the function \( f(n) \) can be confirmed as:

\[
\boxed{f(n) = 25 \cdot \left(-\frac{1}{5}\right)^{n-1}}
\]