Write the series negative 25 minus 22 minus 19 plus ellipsis for 14 terms using summation notation.

Question

(1 point)
Responses

sigma-summation Underscript n equals 1 Overscript 14 EndScripts left-parenthesis negative 28 plus 3 n right-parenthesis
Image with alt text: sigma-summation Underscript n equals 1 Overscript 14 EndScripts left-parenthesis negative 28 plus 3 n right-parenthesis

sigma-summation Underscript n equals 2 Overscript 14 EndScripts left-parenthesis 3 n minus 28 right-parenthesis
Image with alt text: sigma-summation Underscript n equals 2 Overscript 14 EndScripts left-parenthesis 3 n minus 28 right-parenthesis

sigma-summation Underscript n equals 1 Overscript 14 EndScripts left-parenthesis negative 22 minus 3 n right-parenthesis
Image with alt text: sigma-summation Underscript n equals 1 Overscript 14 EndScripts left-parenthesis negative 22 minus 3 n right-parenthesis

sigma-summation Underscript n equals 1 Overscript 14 EndScripts left-parenthesis negative 22 plus 3 n right-parenthesis

Answer 1

To write the series \(-25, -22, -19, \ldots\) for 14 terms using summation notation, we need to determine the general term of the series.

The pattern of the series can be recognized by observing how the terms increase by 3:

• The first term is \(-25\)
• The second term is \(-22\) (which is \(-25 + 3\))
• The third term is \(-19\) (which is \(-22 + 3\))

This gives us a clear formula for the \(n\)-th term of the sequence. The \(n\)-th term can be represented as: \[ a_n = -25 + 3(n-1) \]

This simplifies to: \[ a_n = -28 + 3n \]

Now, we can express the sum of the first 14 terms in summation notation as: \[ \sum_{n=1}^{14} (-28 + 3n) \]

Thus, the correct option in the responses you provided is: \[ \sigma-summation \text{ Underscript n equals 1 Overscript 14 EndScripts } \left( -28 + 3n \right) \]