Problem

Given the following series,
\[
-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}
\]
indicate the corresponding summation notation. Select the correct answer
a) $\sum_{k=1}^{6} \frac{(-1)^{k-1}}{2 k-1}$
b) $\sum_{k=2}^{7} \frac{(-1)^{2 k}+1}{2 k}$
c) $\sum_{k=1}^{6} \frac{(-1)^{k}}{2 k+1}$
d) $\sum_{k=1}^{6} \frac{(-1)^{k-1}}{2 k+1}$

Answer

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Answer

Final Answer: The correct summation notation for the given series is \(\boxed{\sum_{k=1}^{6} \frac{(-1)^{k-1}}{2 k-1}}\).

Steps

Step 1 :Given the following series, \(-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\frac{1}{11}+\frac{1}{13}\)

Step 2 :We need to find the corresponding summation notation. The series is alternating, starting with a negative term, and the denominator of each term is an odd number starting from 3.

Step 3 :This suggests that the correct summation notation should have (-1) raised to a power that alternates between even (for negative terms) and odd (for positive terms) as k increases, and the denominator should be an odd number that starts from 3 when k=1 and increases by 2 as k increases.

Step 4 :This matches with option a) \(\sum_{k=1}^{6} \frac{(-1)^{k-1}}{2 k-1}\).

Step 5 :Final Answer: The correct summation notation for the given series is \(\boxed{\sum_{k=1}^{6} \frac{(-1)^{k-1}}{2 k-1}}\).

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