To find an explicit formula for the sequence defined by the initial term \( a_1 = -4 \) and the recursive formula \( a_n = -a_{n-1} \), we can compute a few terms of the sequence and look for a pattern.
- Given \( a_1 = -4 \)
- Using the recursive formula for \( n=2 \): \[ a_2 = -a_1 = -(-4) = 4 \]
- For \( n=3 \): \[ a_3 = -a_2 = -4 \]
- For \( n=4 \): \[ a_4 = -a_3 = -(-4) = 4 \]
- For \( n=5 \): \[ a_5 = -a_4 = -4 \]
- For \( n=6 \): \[ a_6 = -a_5 = -(-4) = 4 \]
Now we notice a pattern from our calculations:
- For odd \( n \) (1, 3, 5, ...), \( a_n = -4 \)
- For even \( n \) (2, 4, 6, ...), \( a_n = 4 \)
Based on this pattern, we can derive an explicit formula. We can express this using \( n \) as follows:
\[ a_n = \begin{cases} -4 & \text{if } n \text{ is odd} \ 4 & \text{if } n \text{ is even} \end{cases} \]
This can be expressed in a more compact form using the expression: \[ a_n = (-1)^n \cdot 4 \]
This matches our earlier findings:
- When \( n \) is odd, \( (-1)^n = -1 \), so \( a_n = -4 \).
- When \( n \) is even, \( (-1)^n = 1 \), so \( a_n = 4 \).
Thus, the explicit formula for the sequence \( a_n \) is:
\[ \boxed{(-1)^n \cdot 4} \]
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