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PROBLEM

Write the sum using sigma notation:
68+1012+1416+18=n=1A6-8+10-12+14-16+18=\sum_{n=1}^{A} , where
A=B=\begin{array}{l} A= \\ B= \end{array}

STEP 1

1. The sequence alternates between positive and negative terms.
2. The sequence is arithmetic with a common difference.
3. The sequence starts at 6 and ends at 18.

STEP 2

1. Identify the pattern in the sequence.
2. Determine the general formula for the sequence.
3. Write the sequence in sigma notation.
4. Identify the values of AA and BB.

STEP 3

Observe the sequence: 6,8,10,12,14,16,186, -8, 10, -12, 14, -16, 18.
The sequence alternates signs and increases by 2 in absolute value.

STEP 4

Identify the general formula for the sequence.
The sequence can be split into two parts: positive and negative.
Positive terms: 6,10,14,186, 10, 14, 18 which can be written as an=6+4(n1)a_n = 6 + 4(n-1).
Negative terms: 8,12,16-8, -12, -16 which can be written as bn=(8+4(n1))b_n = -(8 + 4(n-1)).

STEP 5

Combine the sequences into a single formula using sigma notation.
The sequence can be expressed as:
n=14(6+4(n1))n=13(8+4(n1))\sum_{n=1}^{4} (6 + 4(n-1)) - \sum_{n=1}^{3} (8 + 4(n-1))

SOLUTION

Identify the values of AA and BB.
The number of terms in the sequence is 7, so A=7A = 7.
The sequence alternates, so BB represents the alternating sign pattern, which is (1)n+1(-1)^{n+1}.
The sigma notation for the sequence is:
n=17(1)n+1(6+2(n1))\sum_{n=1}^{7} (-1)^{n+1} (6 + 2(n-1)) Values:
A=7B=(1)n+1\begin{array}{l} A= 7 \\ B= (-1)^{n+1} \end{array}

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