Questions: -3+5-7+9-11+13-15= sum from n=1 to A of B, where A= B=

-3+5-7+9-11+13-15= sum from n=1 to A of B, where
A= 
B=
Transcript text: -3+5-7+9-11+13-15=\sum_{n=1}^{A} B, where $A=$ $\square$ $B=$ $\square$
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Solution

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Solution Steps

To express the given sum using sigma notation, we need to identify the pattern in the sequence and determine the general term. The sequence alternates between positive and negative numbers and increases by 2 each time. We can use this pattern to find the general term and the number of terms in the sequence.

Step 1: Identify the Sequence

The given sequence is \(-3, 5, -7, 9, -11, 13, -15\). We observe that the terms alternate in sign and increase in absolute value by 2.

Step 2: Determine \( A \)

The number of terms in the sequence is \( A = 7 \).

Step 3: Establish the General Term \( B \)

The general term can be expressed as: \[ B = (-1)^{n+1} (2n - 1) \] This formula captures the alternating signs and the increasing odd numbers.

Final Answer

Thus, we can express the sum using sigma notation as: \[ -3 + 5 - 7 + 9 - 11 + 13 - 15 = \sum_{n=1}^{7} (-1)^{n+1} (2n - 1) \] The values are: \[ A = 7, \quad B = (-1)^{n+1} (2n - 1) \] The final boxed answer is: \[ \boxed{A = 7, \, B = (-1)^{n+1} (2n - 1)} \]

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