To find an equation that describes the given sequence \(-17, 34, -68, \ldots\), we can first observe the pattern in the sequence.
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Recognizing the Pattern:
- The first term (\(n = 1\)) is \(-17\).
- The second term (\(n = 2\)) is \(34\).
- The third term (\(n = 3\)) is \(-68\).
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Finding the Pattern:
- Notice that the absolute values of the terms seem to follow a multiplication pattern:
- \(17 \times 2^0 = 17\)
- \(17 \times 2^1 = 34\)
- \(17 \times 2^2 = 68\)
- Additionally, the signs alternate. The first term is negative, the second is positive, and the third is negative.
- Notice that the absolute values of the terms seem to follow a multiplication pattern:
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General Formula:
- The absolute value of the \(n\)-th term can be formulated as: \[ |a_n| = 17 \cdot 2^{n-1} \]
- The sign can be expressed as \((-1)^n\):
- For \(n = 1\), \((-1)^1 = -1\)
- For \(n = 2\), \((-1)^2 = 1\)
- For \(n = 3\), \((-1)^3 = -1\)
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Putting It All Together:
- Combining both parts, we have: \[ a_n = -17 \cdot 2^{n-1} \cdot (-1)^{n-1} \]
This can also be simplified. Specifically, the effect of the \((-1)^{n-1}\) on the initial negative sign results in the correct alternating signs.
Hence, the equation to describe the sequence is:
\[
a_n = -17 \cdot (-2)^{n-1}
\]
This correctly reflects both the values and alternating signs of the sequence.
Thus the final answer is: \[ a_n = -17 \cdot (-2)^{n-1} \]
3 answers