To find an explicit formula for the sequence defined by the initial term \( a_1 = 15 \) and the recursive formula \( a_n = a_{n-1} + 16 \), we can analyze how the sequence behaves.
Step 1: Calculate the first few terms
Let's compute the first few terms of the sequence:
- \( a_1 = 15 \)
- \( a_2 = a_1 + 16 = 15 + 16 = 31 \)
- \( a_3 = a_2 + 16 = 31 + 16 = 47 \)
- \( a_4 = a_3 + 16 = 47 + 16 = 63 \)
From these calculations, we observe:
\[ \begin{align*} a_1 & = 15 \ a_2 & = 31 \ a_3 & = 47 \ a_4 & = 63 \ \end{align*} \]
Step 2: Identify a pattern
The difference between consecutive terms is always 16. Therefore, the sequence can be described as an arithmetic sequence with a common difference of 16.
Step 3: General form of an arithmetic sequence
The general formula for the \( n \)-th term of an arithmetic sequence can be written as:
\[ a_n = a_1 + (n-1)d \]
where \( d \) is the common difference and \( a_1 \) is the first term.
Step 4: Substitute the known values
In our case:
- \( a_1 = 15 \)
- \( d = 16 \)
Plugging these values into the formula gives:
\[ a_n = 15 + (n-1) \cdot 16 \]
Step 5: Simplify the expression
Now, we simplify this expression:
\[ a_n = 15 + 16(n - 1) \] \[ = 15 + 16n - 16 \] \[ = 16n - 1 \]
Conclusion
Thus, the explicit formula for the sequence \( a_n \) is:
\[ \boxed{16n - 1} \]