Question

vv\begin{cases}d(1)=3\\\\
d(n)=d(n-1)-14
\end{cases}

Answer 1

To solve the recurrence relation given by

\[
\begin{cases}
d(1) = 3 \\
d(n) = d(n-1) - 14 \quad \text{for } n > 1
\end{cases}
\]

we can analyze how the sequence progresses.

1. **Initial Value**: From the first equation, we have:
\[
d(1) = 3
\]

2. **Next Terms**: Using the second part of the definition, we calculate the subsequent terms:
- For \( n = 2 \):
\[
d(2) = d(1) - 14 = 3 - 14 = -11
\]
- For \( n = 3 \):
\[
d(3) = d(2) - 14 = -11 - 14 = -25
\]
- For \( n = 4 \):
\[
d(4) = d(3) - 14 = -25 - 14 = -39
\]
- For \( n = 5 \):
\[
d(5) = d(4) - 14 = -39 - 14 = -53
\]

3. **General Pattern**:
We can see that each term is formed by subtracting 14 from the previous term.

If we track \( d(n) \) over several terms:
- \( d(1) = 3 \)
- \( d(2) = 3 - 14 = 3 - 14 \cdot 1 \)
- \( d(3) = 3 - 14 \cdot 2 \)
- \( d(4) = 3 - 14 \cdot 3 \)
- \( d(5) = 3 - 14 \cdot 4 \)

Thus, we can formulate a general formula:
\[
d(n) = 3 - 14(n-1) \quad \text{for } n \geq 1
\]

4. **Simplifying the General Formula**:
Simplifying \( d(n) \):
\[
d(n) = 3 - 14n + 14 = 17 - 14n
\]

5. **Final Expression**:
Therefore, the explicit formula for \( d(n) \) is:
\[
d(n) = 17 - 14n
\]

This will give you the value of \( d(n) \) for any positive integer \( n \).